Graphically, the derivative of \(f(x)\) at \(x=a\), \(f'(a)\), represents the slope of the line tangent to the graph of \(f(x)\) at the point \(P\ (a,f(a))\).

Finding \(f'(a)\) gives the slope of the tangent line.

The \(y\)-coordinate of the point of tangency can be determined by substituting \(a\) into \(f(x)\) to give \(f(a)\).

Once we have the slope of the tangent line and a point on the tangent line, we have the necessary information for finding the equation of the tangent line.

Determine the equation of the tangent line to the curve \(y=\dfrac{1}{\sqrt{x}}\) at \(x=4\). Express the equation of this line in both standard and \(y=mx+b\) form.

Solution

First, let's determine the point on the graph where the tangent line touches.

Since \(y=\dfrac{1}{\sqrt{x}}\), then when \(x=4\), \(y=\dfrac{1}{\sqrt{4}}= \dfrac{1}{2}\).

Therefore, the point of tangency is \(\left(4, \frac{1}{2}\right)\).

Next, let's use the differentiation rules to find the derivative of \( y\), with respect to \(x\), which will lead us to finding the slope of the tangent line.

Before differentiating, remember that \(y= \dfrac{1}{\sqrt{x}} = x^{-1/2}\).

Now, let's differentiate:

\(\dfrac{dy}{dx} = -\dfrac{1}{2} x^{-\frac{3}{2}} = -\dfrac{1}{2(\sqrt{x})^3}\)

Next, let's substitute the value of \(x\) into \(\dfrac{dy}{dx}\) to determine the slope of the tangent line at \(x=4\).

Determine the equation of the tangent line to the curve \(y=\dfrac{1}{\sqrt{x}}\) at \(x=4\). Express the equation of this line in both standard and \(y=mx+b\) form.

Solution

The notation \(\left. \dfrac{dy}{dx}~\right\vert_{x=4}\) means to substitute \(x=4\) into \(\dfrac{dy}{dx}\).

\( \left. \dfrac{dy}{dx}~\right\vert_{x=4} = -\dfrac{1}{2(\sqrt{4})^3}= -\dfrac{1}{2(2)^3} = -\dfrac{1}{16}\)

Therefore, the slope of the tangent line to \(f(x)=\dfrac{1}{\sqrt{x}}\) at \(x=4\) is \(-\dfrac{1}{16}\).

Finally, we will use the slope of the tangent and the point of tangency to find the equation of the tangent line.

slope \(=-\frac{1}{16}\), point of tangency\( = \left(4, \frac{1}{2}\right)\)

Method 1: Using slope and \(y\)-intercept form.

Substitute the slope and point of tangency into \(y=mx+b\),

Therefore, the equation of the tangent line in \(y=mx+b\) form is \(y=-\dfrac{1}{16}x+\dfrac{3}{4}\).

We could rearrange this formula to express the equation in standard form.

However, let's explore another method for determining the equation of a line.

slope \(=-\frac{1}{16}\), point of tangency\( = \left(4, \frac{1}{2}\right)\)

Method 2: Using slope and point form of a line.

Consider a line with slope \(m\) that passes through the specific point \((x_1, y_1)\).

Using any other general point on the line, \((x,y)\), the slope of the line is then given by \(m=\dfrac{y-y_1}{x-x_1}\).

Rearranging this equation, we have \(y-y_1=m(x-x_1)\) which is the formula for finding the equation of a line given the slope and any point.

We will now use this equation, to determine the equation of the tangent line.

Substitute \( m_{\text{tangent}} = -\dfrac{1}{16}\) and \((x_1, y_1)=\left(4, \dfrac{1}{2}\right)\) into the formula,

Therefore, the equation of the tangent line in standard form is \(x +16y- 12 =0\).

Find the equation of the line tangent to \( y = x^2 + 6x + 9 \) which has a slope of \(6\).

Solution

This question asks for the equation of the tangent line with a slope of \(6\), which implies that there is only one such line.

If we consider the graph of \(y\), this function is a parabola which opens upwards.

Any tangent line drawn to the left of its vertex will have a negative slope, while any tangent line drawn to the right of the vertex will have a positive slope.

Furthermore, as the point of tangency is moved from left to right, the slope of the tangent line continuously increases.

This tells us that every tangent drawn to a point on the graph of \(y=x^2+6x+9\) will have a slope that is different from that at any other point.

Thus, there exists only one point on the parabola where the tangent line has a slope of \(6\).

We already know the slope of the tangent line, but what is unknown is the point of tangency. We can use the slope of the tangent line to find the \(x\)-coordinate of the point of tangency.

Start by differentiating \( y = x^2 + 6x + 9 \).

Find the equation of the line tangent to \( y = x^2 + 6x + 9 \) which has a slope of \(6\).

Solution

Since \(m_{tangent} = 6\), we know that \(\dfrac{dy}{dx}=6\) at the point of tangency.

When \( x = 0 \), \( y = (0)^2 + 6(0) + 9 = 9 \).

Thus, the point of tangency is \( (0, 9) \).

Find the equation of the line tangent to \( y = x^2 + 6x + 9 \) which has a slope of \(6\).

Solution

Recall the slope is \(6\) and the point of tangency is \((0,9)\).

Therefore, the equation of the line tangent line to \( y = x^2 + 6x + 9 \) having slope of \(6\) is \(6x-y+9 = 0\).

At what point(s) does the curve \( y = -\dfrac{1}{3}x^3 + \dfrac{3}{2}x^2\) have a tangent line with slope \(-4\)?

Solution

Since \( y = -\dfrac{1}{3}x^3 + \dfrac{3}{2}x^2 \), then \( \dfrac{dy}{dx} = -x^2 + 3x\). We want to find the values of \(x\) where \( \dfrac{dy}{dx} = -4 \).

Therefore, \(x = 4\) or \(x = -1\).

When \( x = 4\),

\(y = -\dfrac{1}{3}(4)^3 + \dfrac{3}{2}(4)^2 = -\dfrac{64}{3} + \dfrac{48}{2} = \dfrac{8}{3}\)

When \( x = -1\),

\(y = -\dfrac{1}{3}(-1)^3 + \dfrac{3}{2}(-1)^2 = \dfrac{1}{3} + \dfrac{3}{2} = \dfrac{11}{6}\)

Therefore, the tangent lines to the curve \( y = -\frac{1}{3}x^3 + \frac{3}{2}x^2\) with slope \(-4\) touch the curve at points \( \left(4, \frac{8}{3}\right) \) and \( \left(-1, \frac{11}{6}\right) \).

At what point(s) does the curve \( y = -\dfrac{1}{3}x^3 + \dfrac{3}{2}x^2\) have a tangent line with slope \(-4\)?

Solution

In the diagram above, two tangent lines (marked in red) with slope \(-4\) can be drawn to the curve at the indicated points. These lines are parallel to each other. These are the only two points where the tangent line has slope \(-4\).