Alternative Format — Lesson 2: Function Notation

Let's Start Thinking

Functions

Recall

A function is a relation in which each value of the independent variable corresponds to exactly one value of the dependent variable.

We have seen that a vending machine that only accepts exact change operates like a function.

Source: Vending Machine - raddanovic/E+/Getty Images

Every time I input a toonie and press the root beer button, I receive a root beer. If the machine is functioning properly, I won't be surprised by the output after an input of \($2.00\) and a button press because each input has a guaranteed output.

Note: It is possible that multiple inputs can have the same output.

For instance, eight quarters and the cola button will produce the same output as two loonies and the cola button.

The definition of a function only limits what happens for a given input.

Previously, we saw that functions can be described in several ways – tables, graphs, mapping notation, and so on.

In this lesson, we will look at function notation, a new way to represent a function.


Lesson Goals

  • Describe functions using function notation.
  • Analyze linear functions using function notation.
  • Analyze quadratic functions using function notation.

Function Notation


Function Notation Background

There is a limitation in our current way of writing relations. Let's see what this limitation is and how function notation will address the issue.

Consider the relation \(h=-5t^2+20t+1\) describing the height, \(h\), in metres, of a ball thrown in the air where \(t\) is the time in seconds.

The quadratic relation opens downward. It starts when time is zero and ends when the ball reaches a height of zero.

From the equation, we can see that when \(t = 0\), \(h\) is \(1\).

The quadratic relation opens downward. It starts when time is zero and ends when the ball reaches a height of zero. Point labelled at h=1, t=0

  • The \(h\)-intercept is \(1\).
  • Writing this as \(h=1\) is confusing since
    • \(h\) takes on many different values, not just \(h=1\).

      So writing \(h = 1\) without further explanation can be misleading.

    • \(h=1\) corresponds to two different times … \(t=0\) and \(t=4\).

The quadratic relation opens downward. It starts when time is zero and ends when the ball reaches a height of zero. Point labelled at h=1, t=4

So something a little more descriptive would be helpful.

  • This is more clear:
    • \(h_{t=0}=1\), that makes it obvious that we mean when \(t = 0\), \(h = 1\).
    • \(h_{t=4}=1\)
  • This is more precise since the independent and dependent variable values are both included.
  • Function notation works in this way too.

So here's how function notation works.

Consider two functions.

\(h:x\rightarrow \sqrt{x}\)

The square root function turns an input of 9 into an output of 3.

When the input is \(9\), the output is \(3\).

When we show our calculations, we show that we input the number \(9\) into the function by writing \(9\) inside the square root function.

\(\sqrt{9}=3\)

In function notation, we do the same type of thing. 

\(h(9)=\sqrt{9}=3\)

  • \(h(9)\) is the output of \(h\) when the input is \(9\).
  • \(h(9)\) is function notation, not multiplication​.

Let's look at our function \(g\) now.

\(g:x\rightarrow x^3\)

The cubic function turns an input of 2 into an output of 8.

When the input is \(2\), the output is \(8\).

Our calculation shows what we are doing.

\(2^3=8\)

We write \(2\) inside the cubed function, or we can do the same using the name of our function \(g\). 

\(g(2)=2^3=8\)

  • \(g(2)\) is the output of \(g\) when the input is \(2\).
  • \(g(2)\) is function notation, not multiplication​​​​.

Function Notation

So here's a recap.

\(g:x\rightarrow x^3\)

  • In words, the function \(g\) tells us to "take the input and cube it"
  • E.g., Input of 2, \(g(2)\)

    \(g(2)\) is read "g of 2" or "g at 2".

    • \(g(2)=8\)

The cubic function turns an input of 2 into an output of 8.

  • For a general input, \(x\): 

\(g(x)=x^3\)

​​​For any function, \(f:\)

  • when the input is \(x\), the output is ​\(f(x)\).
  • as mapping notation, \(f:x\rightarrow f(x)\).
  • since the output of a function is also \(y\), we can write \(y=f(x)\).

\(f(x)\) is not \(f\) times \(x\) since \(f\) is the name of a function, not a variable.

Rather \(f(x)\) is the output of the function \(f\) when the input is \(x\).


Example 1

For the given function, determine

  1. \(f(4)\)
  2. \(3f(2)-5\)
  3. some value of \(a\) such that \(f(a)=7\)

The function directs 0 to 4, 2 to 6, 3 to 7, 4 to negative 3, 5 to 2, and 7 to 11.

Solution — Part A

Determine \(f(4)\).

\(f(4)\) is the output of the function \(f\) when the input is \(4\).

  • \(4\) from the first oval maps to \(-3\) in the second oval.
  • Therefore, \(f(4)=-3\).

Solution — Part B

Determine \(3f(2)-5\).

While \(f\) is not a number (it is the name of the function), \(f(2)\) is a number.

  • \(f(2)\) is the output of the function \(f\) when the input is \(2\).
  • Therefore, \(f(2)=6\).

Thus,

\(\begin{align*} 3f(2)-5 &= 3(6)-5\\ &= 18-5\\ &= 13 \end{align*}\)

Notice that in this example brackets were used in two ways:

  • In \(f(2)\), the brackets indicate that \(2\) is the input into the function \(f\).
  • In \(3(6)-5\), the brackets indicate that \(3\) and \(6\) must be multiplied together.

Solution — Part C

Determine some value of \(a\) such that \(f(a)=7\).

\(f(a)\) is the output of the function \(f\) when the input is \(a\).

  • Since \(f(a)=7\), we require the input(s), \(a\), that has output \(7\).
    • \(7\) from the second oval corresponds to \(3\) from the first oval.
    • \(7\) is the only output that corresponds to an input of \(3\).
    • Therefore, the only value of \(a\) that satisfies \(f(a)=7\) is \(a=3\).

Example 2

For the given function, determine

  1. \(f(-3)\)
  2. \(f(-1)\)
  3. \(f(2)\)
  4. \(a\) such that \(f(a)=-2\)
This is the graph of a piecewise function. It has a horizontal line at f of x equals negative 4 when x is less than negative 2. The function follows a pattern of f of x equals x minus 1 when x is greater than or equal to negative 2 and less than 2. At x equals 2 there is a dot at the point y equals 3. For x greater than 2 the function equals 4.

Solution — Part A

Determine \(f(-3)\).

\(f(-3)\) is the output of \(f\) when the input is \(-3\) (i.e., \(f(-3)\) is the \(y\)-value that corresponds to \(x=-3\)).

From the graph, when \(x\) is \(-3\), \(y \) is \(-4\).

Therefore, \(f(-3)=-4\).

Highlighted vertical line going through x=negative 3

Solution — Part B

Determine \(f(-1)\).

\(f(-1)\) is the output of \(f\) when the input is \(-1\) (i.e., \(f(-1)\) is the \(y\)-value that corresponds to \(x=-1\)).

From the graph, when \(x\) is \(-1\), \(y\) is \(-2\).

Therefore, \(f(-1)=-2\).

Highlighted vertical line going through x=negative 1

Solution — Part C

Determine \(f(2)\).

\(f(2)\) is the output of \(f\) when the input is \(2\) (i.e., \(f(2)\) is the \(y\)-value that corresponds to \(x=2\)).

From the graph, the two open dots at \(x=2\) are not included in the function; the closed dot is included in the function.

Thus, when \(x\) is \(2\), \(y \) is \(3\).

Therefore, \(f(2)=3\).

Highlighted vertical line going through x=2

Solution — Part D

Determine \(a\) such that \(f(a)=-2\).

\(f(a)\) is the output of \(f\) when the input is \(a\) (i.e., \(f(a)\) is the \(y\)-value that corresponds to \(x=a\)).

The graph has \(y\)-value \(-2\) when \(x\) is \(-1\).

Since \(f(-1)=-2\), \(a=-1\).

Highlighted horizontal line going through y=negative 2

Check Your Understanding 1

Question

For the given function, determine:

  1. \(f(3)\)
  2. \(f(3)+f(8)\)
  3. \(f(3+8)\)
  4. \(a\) such that \(f(a)=2\)
\(x\) \(f(x)\)
\(-7\) \(10\)
\(-6\) \(4\)
\(3\) \(7\)
\(7\) \(-9\)
\(8\) \(2\)
\(11\) \(11\)

Answer

  1. \(7\)
  2. \(9\)
  3. \(11\)
  4. \(8\)

Feedback

  1. When the input into \(f\) is \(3\), the output is \(7\). Therefore, \(f(3) = 7\).
  2. Since \(f(3) = 7\) and \(f(8) = 2\),

    \(\begin{align*} f(3) + f(8) &= 7 + 2 \\ &= 9 \end{align*}\)

  3. \(f(3+8) = f(11) = 11\)
  4. \(f(a)\) is the output of \(f\) when the input is \(a\). Therefore, since \(f(8) = 2\), \(a =8\).

Function Notation

We will now look at how function notation is used when the function is defined algebraically.

Recall

\(f(x)\) is the output of the function \(f\) when the input is \(x\).

Although, \(f(x)\) looks like multiplication, we can remember that it's not, since \(f\) is the name of a function, not a number.

Let's try an example.

Example 3

Given \(f(x)=4x+3\) and \(g(x)=3x^2-x+1\), determine

  1. \(f(6)\)
  2. \(g(-3)\)
  3. \(g(5-2)\)
  4. \(2f(5)-7\)

Solution — Part A

Determine \(f(6)\):

  • The input value, or \(x\)-value, is \(6\).
  • The output value is \(f(6)\).

To calculate \(f(6)\), start by writing out the function.

\(f(x) = 4 x+3\)

Then replace each \(x\) with \(6\) and evaluate.

\(\begin{align*} f(6) &= 4(6)+3 \\ f(6) &= 27 \end{align*}\)

Therefore, \(f(6)=27\).

Solution — Part B

Recall part b): Determine \(g(-3)\).

\(\begin{align*}g(x) &=3 x^{2}-x+1 \\ g(-3) &=3(-3)^{2}-(-3)+1 \\ g(-3) &=27+3+1 \\ g(-3) &=31 \end{align*}\)

Therefore, \(g(-3) = 31\).

Solution — Part C

Recall part c): Determine \(g(5-2)\)

\(\begin{align*} g(5-2) &=g(3) \\ &=3(3)^{2}-3+1 \\ &=25 \end{align*}\)

Therefore, \(g(5-2) = 25\).

Before we move on, let's check if the distributive property is true for function notation.

So in other words, is \(g(5-2)\) the same as \(g(5)-g(2)\)?

\(\begin{align*} g(5)-g(2) &=3(5)^{2}-5+1-\left(3(2)^{2}-2+1\right) \\ &=71-11 \\ &=60 \\ & \neq 25 \end{align*}\)

\(g(5)-g(2)\) is not the same as \(g(5-2)\).

In general, \(g(a)+g(b) \ne g(a+b)\) for \(a,b\in\mathbb{R}\).

Solution — Part D

Recall part d): Determine \(2f(5)-7\)

Note that:

  • \(f(5)\) is the output of the function, \(f\), when the input is \(5\). 
  • First calculate \(f(5)\), then take that answer and multiply it by \(2\) and subtract \(7\).

    \(\begin{align*} 2 f(5)-7 &=2[4(5)+3]-7 \\ &=2(23)-7 \\ &=39 \end{align*}\)

Therefore, \(2f(5)-7 = 39\).


Example 4

Given \(f(x)=\sqrt{x+5}\), determine \(f(31)\).

Solution

Inputting \(x=31\) into the function \(f(x)\):

\(\begin{align*} f(x) &= \sqrt{x+5}\\ f(31) &= \sqrt{31+5}\\ &= \sqrt{36}\\ &= 6 \end{align*}\)


Check Your Understanding 2

Question — Version A

\(f(x)= x^2 - 4x\), determine \(f(4)\).

Answer — Version A

The value of \(f(4)\) is \(0\).

Feedback — Version A

To evaluate \(f(4)\), substitute \(4=2\) into \(f(x)= x^2 - 4x\)

\(\begin{align*} f(x) &= x^2 - 4x \\ f(4) &= 4^2 - 4(4) \\ &= 0  \end{align*}\)

Therefore, the value of \(f(4)\) is \(0\).

Question — Version B

\(g(x)= \dfrac{30}{8 - x}\), determine \(f(2)\).

Answer — Version B

The value of \(g(2)\) is \(5\).

Feedback — Version B

To evaluate \(g(2)\), substitute \(x=2\) into \(g(x)= \dfrac{30}{8 - x}\)

\(\begin{align*} g(x) &= \dfrac{30}{8-x} \\ g(2) &= \dfrac{30}{8 - (2)} \\ &= 5  \end{align*}\)

Therefore, the value of \(g(2)\) is \(5\).


Example 5

Given \(h(x)=3x^2-2x+6\), determine \(h(a-3)\) in simplified form.

Solution

Notice that in this example we have an algebraic input into the function.

open bracket a minus 3 closed bracket produces an output of h(a minus 3)

We proceed in the usual way, replacing each \(x\) in the function notation with \(a-3\):

\(\begin{align*} h(\class{bg1}{x})&=3\class{bg1}{x}^2-2\class{bg1}{x}+6\\ h(\class{bg1}{a-3}) &= 3 (\class{bg1}{a-3})^2-2(\class{bg1}{a-3})+6\\ &= 3(a-3)(a-3)-2a+6+6\\ &= 3(a^2-6a+9)-2a+12\\ &= 3a^2-18a+27-2a+12\\ &= 3a^2-20a+39 \end{align*}\)

Therefore, \(h(a-3)= 3a^2-20a+39\). 


Check Your Understanding 3

Question

Given \(h(x)= -2x^2 + 3x - 2\), determine \(h(p - 2)\).

Answer

\(h(p-2) = -2p^2 + 11p - 16\)

Feedback

To determine ‌\(h(p - 2)\), evaluate ‌\(h(x)\) at ‌\(x= p - 2\):

\(\begin{align*} h(x)&= -2x^2 + 3x - 2\\ h(p-2)&= -2(p-2)^2 + 3(p-2) - 2\\ &=-2(p^2 - 4p + 4) + 3p - 6 -2\\ &=-2p^2 + 8p - 8 + 3p - 8 \\ &= -2p^2 + 11p - 16 \end{align*}\)


Function Notation and Linear Functions


Let's turn our attention to analyzing linear functions using function notation.

Example 6

For the function \(C(F)=\dfrac{5}{9}\left(F-32\right)\) , where \(C(F)\) is the Celsius temperature, in \(^\circ\)C, that is equivalent to the Fahrenheit temperature, \(F \), in \(^\circ\)F:

  1. Determine \(C(77)\) and interpret its meaning.
  2. Determine \(F\) where \(C(F)=10\) and interpret its meaning.
  3. Determine the temperature \(x\) such that \(C(x)=x\) and interpret its meaning.

A thermometer with degrees Celsius on the left and the equivalent degrees Fahrenheit on the right.

Source: Thermometer - Milena_Vuckovic/iStock/Getty Images

Solution — Part A

To determine \(C(77)\), we evaluate \(C(F)=\dfrac{5}{9}\left(F-32\right)\) at \(F=77\):

\(\begin{align*} C(F) &= \frac{5}{9}(F-32)\\ C(77) &= \frac{5}{9}(77-32)\\ &= \frac{5}{9}(45)\\ &= 25 \end{align*}\)

Therefore, \(C(77)=25\).

This means that \(25^\circ\)C is equivalent to \(77^\circ\)F.

77 degrees Fahrenheit is equivalent to 25 degrees Celsius.

Solution — Part B

Determine \(F\) when \(C(F)=10\) and interpret its meaning. 

\(\begin{align*} C(F) &= \frac{5}{9}(F-32)\\ 10 &= \frac{5}{9}(F-32) & \text{Substitute } C(F)=10 \\ 18 &= F-32 & \text{Multiply both sides by } \frac{9}{5} \\ F &= 50 \end{align*}\)

Therefore, \(F=50\).

This means that \(50^\circ\)F is equivalent to \(10^\circ\)C since \(C(50)=10\).

50 degrees Fahrenheit is equivalent to 10 degrees Celsius.

Solution — Part C

Determine the temperature \(x\) such that \(C(x)=x\) and interpret its meaning.

\(\begin{align*} C(F) &= \frac{5}{9}(F-32)\\ C(x) &= \frac{5}{9}(x-32) & \text{Input } x \text { into } C(F)\\ x &= \frac{5}{9}(x-32) & \text{Substitute } C(x)=x\\ 9x &= 5(x-32) & \text{Multiply both sides by } 9 \\ 9x &= 5x-160\\ 4x&=-160\\ x&=-40 \end{align*}\)

Therefore, \(x=-40\).

This means that the Celsius and Fahrenheit scales are equivalent at
 \(-40^\circ\)C\(=-40^\circ\)F  since \(C(-40)=-40\).

Negative 40 degrees Fahrenheit is equivalent to negative 40 degrees Celsius.

Source: Thermometer - Milena_Vuckovic/iStock/Getty Images

Example 7

Jason delivers newspapers five days a week. Each day he delivers to \(127\) houses. He earns \(13\) cents per paper that he delivers. Determine the

  1. mapping notation that describes this situation.
  2. function notation that describes this situation.
  3. amount Jason earns in two weeks.
  4. number of weeks it will take to earn \($1000\).

Solution — Part A

Let \(E(p)\) be Jason's earnings, in dollars, when he sells \(p\) papers .

Since Jason earns \(13\) cents per paper, or \($0.13\) per paper:

\(E:p \rightarrow 0.13p\)

Solution — Part B

Since \(E:p \rightarrow 0.13p\):

\(E(p)=0.13p\)

Solution — Part C

In two weeks, Jason delivers \(127\) papers each day for \(10\) days for a total of \(1270\) total papers delivered.

His earnings are calculated as follows:

\(\begin{align*} E(p)&=0.13p\\ E(1270)&=0.13(1270)\\ &= 165.1 \end{align*}\)

Therefore, Jason earns \($165.10\) in a two-week period.

Solution — Part D

We can determine how many papers Jason needs to deliver to earn \($1000\) by finding \(p\) such that \(E(p)=1000\):

\(\begin{align*} E(p)&=0.13p\\ 1000&=0.13p\\ p &\approx 7692.3 \end{align*}\)

This value needs to be rounded up to ensure that the earnings are at least \($1000\).

Thus, Jason needs to deliver at least \(7693\) papers.

  • Since Jason delivers \(127\) papers per day, this corresponds to \(7693 \div 127 \approx 61\) days.
  • Since Jason delivers \(5\) days per week, this corresponds to \(61 \div 5 =12.2\) weeks.

Therefore, Jason will have earned a total of \($1000\) during the \(13\)th week.

To answer part d), we could have used our result from part c). Can you see how we could have done that?

Hint: You can use part c) to determine Jason's weekly rate of pay.

Example 8

Determine the linear function that contains the points \(A(5,-3)\) and \(B(-7,7)\). Express your answer using function notation.

 Solution

A graph of the line is:

The graph of a linear function passing through the point (5, -3) and (-7, 7).

Recall the steps used previously to determine the equation of a line given two points on the line.

Recall

The equation of a non-vertical line is of the form \(y=mx+b\) or \(f(x)=mx+b\) where \(m\) is the slope and \(b\) is the \(y\)-intercept.

To determine the equation of a line given two points, \((x_1, y_1)\) and \((x_2, y_2)\), on the line,

  1. Determine the slope, \(m=\dfrac{y_2-y_1}{x_2-x_1}\).
  2. Determine the \(y\)-intercept, \(b\).

Step 1: Determine the slope

\(\begin{align*} m &= \dfrac{y_2-y_1}{x_2-x_1}\\ &= \dfrac{7+3}{-7-5}\\ &= \dfrac{10}{-12}\\ &= \dfrac{-5}{6} \end{align*}\)

Step 2: Determine the \(y\)-intercept

The equation is of the form \(f(x)=\dfrac{-5}{6}x+b\).

Either \(A(5,-3)\) or \(B(-7,7)\) can be used to determine \(b\). Using \(A(5,-3)\), we note that \(f(5)=-3\):

\(\begin{align*} f(x)&=\dfrac{-5}{6}x+b\\ f(5)&=\dfrac{-5}{6}(5)+b\\ -3 &= \dfrac{-5}{6}(5)+b\\ -3 &= \dfrac{-25}{6}+b\\ b &=-3 + \dfrac{25}{6}\\ b &= \dfrac{-18}{6} + \dfrac{25}{6}\\ b &= \dfrac{7}{6} \end{align*}\)

Therefore, the function \(f(x)=\dfrac{-5}{6}x+\dfrac{7}{6}\) is the linear function that contains \(A(5,-3)\) and \(B(-7,7)\).


Check Your Understanding 4

Question

Determine the linear function \(f\) that contains the points \(A(-2,7)\) and \(B(-6,-1)\). Express your answer as an equation using function notation.

Answer

The function \(f(x)=2x+11\) is the linear function that contains \(A(-2,7)\) and \(B(-6,-1)\).

Feeback

To determine the equation of a line given two points on the line, determine the slope, and the \(y\)-intercept.

Step 1: Determine the slope

\(\begin{align*} m &= \dfrac{y_2-y_1}{x_2-x_1}\\ &= \dfrac{-1-7}{-6+2}\\ &= 2 \end{align*}\)

Step 2: Determine the \(y\)-intercept

The equation is of the form \(f(x)=2x+b\).

Either \(A(-2,7)\) or \(B(-6,-1)\) can be used to determine \(b\). Using \(A(-2,7)\), we note that \(f(-2)=7\):

\(\begin{align*} f(x)&=2x+b\\ f(-2) &=2(-2) + b\\ 7 &= -4 +b\\ b &=7+4\\ b &= 11 \end{align*}\)

Therefore, the function \(f(x)=2x+11\) is the linear function that contains \(A(-2,7)\) and \(B(-6,-1)\).


Function Notation and Quadratic Functions


Let's move on to analyzing quadratic functions using function notation.

Example 9

A company sells mugs. The profit function for this company is \(P(x)=-2x^2+12x-10\) where \(P(x)\) is the profit (in thousands of dollars) when the mugs are sold for \(x\) dollars.

Using three different methods, determine the maximum profit.

Solution

To start, let's get a sense of what the graph looks like.

\(\begin{align*} P(0) &=-2(0)^2+12(0)-10\\ &= -10 \end{align*}\)

  • giving mugs away for free
  • no income; loss of \($10~000\)
  • negative profit

So the graph has \(P\)-intercept of \(-10\).

The point (0, negative 10) is on the function, P of x.

Because the coefficient of \(x^2\) is negative, \(-2\), the parabola opens down.

We'll sketch the parabola like this.

The parabola opens downward.

The maximum profit occurs at the vertex.

We don't actually know yet if the vertex is above or below the \(x\)-axis. Hopefully, for this company though, the vertex is above the \(x\)-axis because the maximum profit occurs at the vertex.

Three Methods:

  1. Change \(P(x)\) to factored form
    • From there, we'll be able to use the zeros to help us find the maximum.
  2. Change \(P(x)\) to vertex form
  3. Use the vertex formula
    • You may not remember this formula from previous lessons, nor should you really, as we will show how it can be quickly found from the quadratic formula.

Method 1: Change \(P(x)\) to factored form

Recall

Types of Factoring:

  1. Common factoring
  2. Factoring \(ax^2+bx+c\)
  3. Factoring difference of squares
  4. Factoring perfect squares

There are other types, but these are the four most common types of factoring.

Notice that \(P(x)=-2x^2+12x-10\) is of the form \(ax^2 + bx + c\).

Let's not jump straight away to the second type of factoring because we can also notice that there is a common factor of \(-2\).

\(\begin{align*} P(x) &=-2 x^{2}+12 x-10 \\ &=-2\left(x^{2}-6 x+5\right) \end{align*}\)

So whenever possible, remove common factors before applying the other types of factoring.

The remaining factor is of the form \(x^2 + bx + c\), which will lead us now to the second type of factoring.

\(P(x) =-2\left(x^{2}-6 x+5\right) \)

We require two integers that multiply to \(5\), and add to \(-6\).

The integers are \(-1\) and \(-5\).

\( P(x) =-2(x-1)(x-5) \)

Since we remove the common factor first, the expression we are factoring has an \(x^2\) coefficient of \(1\).

As a result, we avoid decomposition.

\(\begin{align*} P(x) &=-2 x^{2}+12 x-10 \\ &=-2\left(x^{2}-6 x+5\right) \\ &=-2(x-1)(x-5) \end{align*}\)

From this equation, we know that the zeros are \(1\) and \(5\).

Let's add this to our graph.

So from earlier, we know that the \(P\)-intercept is \(-10\). We now can add in the zeros are at \(1\) and \(5\).

These are the two break even points, the prices for which the profit is \(0\).

Because a parabola is symmetric, we know that the axis of symmetry must lie here at \(x = 3\).

So the maximum occurs at the vertex. So that means the maximum must be \(P(3)\).

We'll use the standard form expression to calculate \(P(3)\), though we could have used the factored form expression.

\(\begin{align*} P(3) &\;= -2(3)^2+12(3)-10\\ &= 8 \end{align*}\)

Some arithmetic yields \(P(3) = 8\). And since \(P(x)\) is the profit in thousands of dollars, we find that the maximum profit is \($8,000\).

Method 2: Change \(P(x)\) to vertex form

Steps to complete the square

\(P(x)=-2x^2+12x-10\)

  1. Common factor \(a\) from first two terms.

    \(P(x) =-2\left(x^{2}-6 x\right)-10 \)

  2. Add and subtract the value that completes the square. To determine this value, take half of the \(x\)-coefficient and square it.

    \(P(x) =-2\left(x^{2}-6 x+9-9\right)-10 \)

  3. Factor the perfect square

    \(P(x) =-2\left[(x-3)^{2}-9\right]-10 \)

  4. Distribute \(a\).

    \(P(x) =-2(x-3)^{2}+18-10 \)

  5. Collect like terms.

    \(P(x) =-2(x-3)^{2}+8 \)

Since this is now in vertex form, we can see that the vertex is \((3,8)\).

The maximum profit occurs at the vertex. Thus the maximum profit is \($8000\).

Method 3: Use the vertex formula

Recall

The zeros of \(y=ax^2+bx+c\) are 

\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)

Zeros are

\(\dfrac{-b}{2a} - \dfrac{\sqrt{b^2-4ac}}{2a}\)

and

\(\dfrac{-b}{2a} + \dfrac{\sqrt{b^2-4ac}}{2a}\)

Since the axis is halfway between the two zeros, it must then be that the axis of symmetry is \(x=\dfrac{-b}{2a}\).

So we have a formula for the axis of symmetry.

Since the vertex lies on the axis of symmetry, we can find the \(x\)-coordinate of the vertex.

Substituting our values into this formula,

\(\begin{align*} x_{\text { vertex }} &=\frac{-b}{2 a} \\ &=\frac{-12}{-2(2)} \\ &=3 \end{align*}\)

Then to determine the maximum, we evaluate \(P(3)\).

\(\begin{align*} P(3) &= -2(3)^2+12(3)-10\\ &= 8 \end{align*}\)

So we find the same graph as before, and we arrive at the same maximum profit.

The maximum profit is \($8000\).


Example 10

Determine the quadratic function with:

  1. Zeros \(5\) and \(-3\) and \(y\)-intercept \(-4\).
  2. Zeros \(x=5 \pm 2 \sqrt{3}\) that passes through \((6,-1)\).

Solution — Part A

Determine the quadratic function with zeros \(5\) and \(-3\) and \(y\)-intercept \(-4\).

We start with a graph.

The quadratic function must pass through the points (negative 3, 0), (0, negative 4), and (5, 0), resulting in a parabola that opens up.

Since the zeros are \(5\) and \(-3\), the function must be of the form \(f(x)=a(x-5)(x+3)\).

Since the \(y\)-intercept is \(-4\), we note that \(f(0)=-4\):

\(\begin{align*} f(x) &= a(x-5)(x+3)\\ f(0)&= a(0-5)(0+3)\\ -4 &= a(-5)(3)\\ -4 &= -15a\\ a &= \dfrac{4}{15} \end{align*}\)

Therefore, \(f(x)=\dfrac{4}{15}(x-5)(x+3)\) is the quadratic function with zeros \(5\) and \(-3\) and \(y\)-intercept \(-4\).

Solution — Part B

Determine the quadratic function with zeros \(x=5 \pm 2 \sqrt{3}\) that passes through \((6,-1)\).

We start with a graph.

The function opens upwards.

Since the zeros are \(5+2\sqrt{3}\) and \(5-2\sqrt{3}\), the function must be of the form

\(\begin{align*} f(x)&=a[x-(5+2\sqrt{3})][x-(5-2\sqrt{3})]\\ &=a(x-5-2\sqrt{3})(x-5+2\sqrt{3}) \end{align*}\)

Since \((6,-1)\) is on the curve, we note that \(f(6)=-1\):

\(\begin{align*} f(x)&=a(x-5-2\sqrt{3})(x-5+2\sqrt{3})\\ f(6)&= a(6-5-2\sqrt{3})(6-5+2\sqrt{3})\\ -1 &= a(1-2\sqrt{3})(1+2\sqrt{3})\\ \end{align*}\)

Using the distributive property:

\(\begin{align*} -1 &= a(1-2\sqrt{3})(1+2\sqrt{3})\\ -1 &= a(1+2\sqrt{3}-2\sqrt{3}-4\sqrt{9})\\ -1 &= a(1+2\sqrt{3}-2\sqrt{3}-12)\\ -1 &= a(1-12)\\ -1 &= -11a\\ a &= \dfrac{1}{11} \end{align*}\)

Therefore, \(f(x)=\dfrac{1}{11}(x-5-2\sqrt{3})(x-5+2\sqrt{3})\) is the quadratic function with zeros \(5\pm2\sqrt{3}\) that passes through \((6,-1)\).


Check Your Understanding 5

Question

Determine the quadratic function \(f(x)\) with zeros \(1\pm 5\sqrt{6}\) that passes through \((-2, -4)\).

Answer

 

Feedback

Since the zeros are \(1 + 5\sqrt{6}\) and \(1 - 5\sqrt{6}\), the function must be of the form

\(\begin{align*} f(x)&= a(x - (1 + 5 \sqrt{6}))(x - (1 - 5 \sqrt{6})) \\ &= a(x - 1 - 5 \sqrt{6})(x - 1 + 5 \sqrt{6}) \end{align*}\)

Since \((-2, -4)\) is on the curve, we note that \(f(-2)= -4\): 

\(\begin{align*} f(x)&= a(\class{hl1}{x} - 1 - 5 \sqrt{6})(\class{hl1}{x} - 1 + 5 \sqrt{6})\\ f(-2)&= a(\class{hl1}{-2} - 1 - 5 \sqrt{6})(\class{hl1}{-2} - 1 + 5 \sqrt{6})\\ -4 &= a(-3 - 5 \sqrt{6})(-3 + 5 \sqrt{6})\\ \end{align*}\)

Using the distributive property:

\(\begin{align*} -4 &= a(-3 - 5 \sqrt{6})(-3 + 5 \sqrt{6}) \\ -4 &= a(9 + 15\sqrt{6} - 15\sqrt{6} - 150)\\ -4 &= a(9 - 150) \\ -4 &= a(-141) \\ a &= \dfrac{4}{141} \end{align*}\)

Therefore, \(f(x)= \dfrac{4}{141}(x - 1 - 5 \sqrt{6})(x - 1 + 5 \sqrt{6})\) is the quadratic function with zeros \(1\pm 5\sqrt{6}\) that passes through \((-2, -4)\).


Example 11

Determine the intersection of the line \(f(x)=2x+3\) and the parabola \(g(x)=x^2-2x-18\).

Solution

Here is a graph of this situation:

The functions f of x and g of x could intersect at 2 points.

The \(x\)-values of the points of intersection must satisfy \(f(x)=g(x)\) since they are the values of \(x\) that have the same \(y\)-values:

\(\begin{align*} f(x) &= g(x)\\ 2x+3 &= x^2-2x-18 \end{align*}\)

Since this is a quadratic equation, we write the equation in the form \(ax^2+bx+c=0\):

\(\begin{align*} 2x+3 &= x^2-2x-18\\ 3 &= x^2-4x-18\\ 0 &= x^2-4x-21\\ \end{align*}\)

To solve \(x^2-4x-21=0\), we can either factor or use the quadratic formula. Let's factor.

  • We require two integers that multiply to \(-21\) and add to \(-4\).
  • The integers are \(-7\) and \(3\).

Therefore, factored form of \(x^2-4x-21=0\) is

\((x-7)(x+3)=0\)

The solutions of this equation are \(x=7\) and \(x=-3\).

Therefore, the points of intersection occur at \(x=7\) and \(x=-3\).

The \(y\)-coordinates of the points of intersection can be determined by substituting \(x=7\) and \(x=-3\) into either \(f(x)\) or \(g(x)\). Since \(f(x)\) is a simpler function, we use it:

\(\begin{align*} f(x) &= 2x+3\\ f(7) &= 2(7)+3\\ &= 17 \end{align*}\)

\(\begin{align*} f(x) &= 2x+3\\ f(-3) &= 2(-3)+3\\ &= -3 \end{align*}\)

Therefore, the points of intersection are \((7,17)\) and \((-3,-3)\).

Checking Our Work

When calculating the \(y\)-coordinates of our points of intersection, we used \(y=f(x)\). While not a full check of our work, let's verify that using \(y=g(x)\) yields the same result:

\(\begin{align*} g(x) &= x^2 - 2x -18\\ g(7) &= (7)^2 - 2(7) -18\\ &= 49-14-18\\ &= 17 \end{align*}\)

\(\begin{align*} g(x) &= x^2 - 2x -18\\ g(-3) &= (-3)^2-2(-3)-18\\ &= 9+6-18\\ &= -3 \end{align*}\)

As expected, \((7,17)\) and \((-3,-3)\) are points on \(y=g(x)\) as well as on \(y=f(x)\).


Check Your Understanding 6

Question — Version A

Without graphing, find the only point of intersection of the following functions.

\(f(x)=6x +2\)

\(g(x)=x^2 + 4x +3\)

Answer — Version A

The functions \(f(x)=6x +2\) and \(g(x)=x^2 + 4x +3\) intersect at the point \((1, 8)\).

Feedback — Version A

\(\begin{align*} f(x) &= 6x+2 \tag{1} \\ g(x) &= x^2+4x+3 \tag{2} \end{align*}\)

At the point of intersection,  \(f(x) = g(x) \).

\[\begin{align*} 6x+2 &= x^2+4x+3 \\ 0&=x^2+4x+3 - (6x+2) \\ 0&=x^2-2x+1 \\ 0&= (x-1)^2 \end{align*}\]

So \(x=1\). This is the \(x\)-coordinate of the only point of intersection.

To find the \(y\)-coordinate, substitute \(x=1\) into equation \((1)\).

\[\begin{align*} f(x) &= 6x+2 \\ f(1)&= 6(1)+2 \\ &= 8 \end{align*}\]

Therefore, the functions \(f(x)=6x +2\) and \(g(x)=x^2 + 4x +3\) intersect at the point \((1, 8)\).

Question — Version B

Without graphing, find the two points of intersection of the following functions.

\(f(x)=9x -4\)

\(g(x)=2x^2 - 5x -40\)

Answer — Version B

The functions \(f(x)=9x -4\) and \(g(x)=2x^2 - 5x -40\) intersect at the points \((9, 77)\) and \((-2, -22)\).

Feedback — Version B

\(\begin{align*} f(x) &=9x -4 \tag{1} \\ g(x) &=2x^2 - 5x -40 \tag{2} \end{align*}\)

At the point of intersection,  \(f(x) = g(x) \).

\[\begin{align*} 9x - 4 &=2x^2 - 5x -40 \\ 0&=2x^2 - 5x -40 - (9x - 4) \\ 0&=2x^2 - 14x - 36 \\ 0&= 2(x+2)(x-9) \end{align*}\]

So \(x= 9\) or \(x=-2\). These are the \(x\)-coordinates of the points of intersection.

To find the \(y\)-coordinate, substitute \(x= 9\) and \(x= -2\) into equation \((1)\).

\(\begin{align*} f(x) &= 9x -4 \\ f(9)&= 9(9) -4 \\ &= 77 \end{align*}\)

\(\begin{align*} f(x) &= 9x -4 \\ f(-2) &= 9(-2) -4 \\ &= -22 \end{align*}\)

Therefore, the functions \(f(x)=9x -4\) and \(g(x)=2x^2 - 5x -40\) intersect at the points \((9, 77)\) and \((-2, -22)\).


Example 12

For what real values of \(k\) will \(f(x)=3x+k\) and \(g(x)=3x^2-2x+4\) have two intersection points?

Solution

If an intersection point exists, its \(x\)-coordinate must satisfy \(f(x)=g(x)\):

\(\begin{align*} f(x) &= g(x)\\ 3x+k &= 3x^2-2x+4\\ 3x^2-5x+4-k&=0 \end{align*}\)

Recall

A quadratic equation of the form \(ax^2+bx+c=0\) has

  • Two distinct real solutions when \(b^2-4ac \gt 0\).
  • One real solution when \(b^2-4ac=0\).
  • No real solutions when \(b^2-4ac \lt 0\).

\(b^2-4ac\) is called the discriminant.

For \(y=f(x)\) and \(y=g(x)\) to have two intersection points, we require the equation \(3x^2-5x+4-k=0 \) to have two distinct real solutions. This happens when \(b^2-4ac \gt 0\):

\(\begin{align*} b^2-4ac &\gt 0\\ (-5)^2-4(3)(4-k) &\gt 0\\ 25-12(4-k) &\gt 0\\ 25-48 +12k &\gt 0\\ -23 + 12k &\gt 0\\ 12k &\gt 23\\ k &\gt \dfrac{23}{12} \end{align*}\)

Therefore, \(f(x)=3x+k\) and \(g(x)=3x^2-2x+4\) have two intersection points for all real numbers \(k \gt \dfrac{23}{12}\).

Recall

An inequality is solved in the same way as an equation, except for one additional rule: if you multiply or divide both sides of an inequality by a negative value, reverse the inequality sign.

While not used in this Courseware, it should also be noted that the inequality sign needs to be reversed when taking the reciprical of both sides too.


Wrap-Up


Lesson Summary

  • Function notation was introduced: \(f(x)\) is the output of the function \(f\) when the input is \(x\).
    • \(f\) is not a number; it is the name of a function.
    • \(f(x)\) is a number; it is the output of the function when the input is \(x\).
    • Since both \(y\) and \(f(x)\) represent the outputs of a function, \(y=f(x)\).
  • Function notation was applied in various contexts.
    • Given the output of a function, we found the corresponding input by solving an equation.
    • Given two points on a line, we determined the linear function by first finding the slope and then the \(y\)-intercept.
    • Given two zeros and a third point on a parabola, we determined the quadratic function. This was also done when the zeros were irrational.

Take It With You

We have seen that the input into a function can be a given number (e.g., Determine \(f(4)\)).

We have also seen that the input into a function can be an algebraic expression (e.g., Determine \(f(a-5)\)).

The input into a function can also involve another function. Can you show that if \(f(x)=3x-5\) and \(g(x)=\sqrt{x+4}\), then \(f(g(5))=4\)?