2. Polynomial Functions
  3. Graphs of Polynomial Functions in Factored Form

Partial Solutions


  1. There is no solution provided for this question.
    1. A quadratic function with negative leading coefficient will have end behaviour \( y \rightarrow -\infty \) as \( x \rightarrow \pm \infty \). Since the zero at \( x = -5 \) has multiplicity \(2\), it will be the only zero and a turning point will occur at \( x = -5 \).
      Quadratic with turning point at (-5, 0), touching from above.
    2. A 5th degree function with a positive leading coefficient will have end behaviour \( y \rightarrow -\infty \) as \( x \rightarrow -\infty \) and \( y \rightarrow \infty \) as \( x \rightarrow \infty \). The zero at the origin has order \(2\), so a turning point occurs at \( x = 0 \). The zero at \( x = 3 \) has order \(3\) so a point of inflection occurs at \( x = 3 \).
      Quintic with turning point at (0, 0), touching from below; inflection point at (3, 0).
    3. A quartic with positive leading coefficient will have end behaviour \( y \rightarrow \infty \) as \( x \rightarrow \pm \infty \). Since both zeros are of order \(2\), a turning point will occur at each.
      Quartic with turning points at (0, 0) and (3, 0), touching from above.
    4. A cubic with negative leading coefficient will have end behaviour \( y \rightarrow \infty \) as \( x \rightarrow -\infty \) and \( y \rightarrow -\infty \) as \( x \rightarrow \infty \). Since it only has one real zero, at \( x = 4 \), with two non-real zeros, the real zero must be multiplicity \(1\). Thus the curve will pass directly through the \( x \)-axis at \( x = 4 \). Since the function is cubic, the curve may have two turning points, or no turning point, with a shape similar to \( y = x^3 \).
      Negative cubic with single zero at x = 4, two turning points.
    5. A quintic function with positive leading coefficient has end behaviour \( y \rightarrow -\infty \) as \( x \rightarrow -\infty \) and \( y \rightarrow \infty \) as \( x \rightarrow \infty \). The graph passes directly through \( x = -2 \) since the zero has multiplicity \(1\), and has turning point at \( x = 1 \) since this zero has multiplicity \(4\).
      Positive quintic with turning point tangent to x-axis at x = 1, touching from above, and passing through x-axis at x = -2
  3. There is no solution provided for this question.
    1. The function is a cubic with negative leading coefficient so \( y \rightarrow -\infty \) as \( x \rightarrow -\infty \) and \( y \rightarrow \infty \) as \( x \rightarrow \infty \). There are zeros at \( x = -2, 0, \frac{5}{2} \), all of multiplicity \( 1 \). So the graph will pass directly through the \( x \)-axis at each zero. Thus, a possible graph is:
      Sketch of y = -x(x + 2)(2x - 5)
    2. The function is a quartic with a positive leading coefficient, so \( y \rightarrow \infty \) as \( x \rightarrow \pm \infty \). There are zeros of order \(2\) at \( x = -3, 2 \), which produce turning points tangent to the \( x \)-axis from above. Thus, a possible graph is:
      Sketch of y = 2(x - 2)^2(x + 3)^2
    3. The function is a quartic with a negative leading coefficient of \( -0.5 \) ( since \( g(x) = -0.5(x - 3)(x + 1)^3 \)) so \( y \rightarrow -\infty \) as \( x \rightarrow \pm \infty \). There is a zero of order \(1\) at \( x = 3 \), so the graph passes through the \( x \)-axis here. There is a zero of order \(3\) at \( x = -1 \), so the graph will linger close to the \( x \)-axis as it passes through and a point of inflection occurs at this zero. Thus, a possible graph is:
      Sketch of y = -0.5(3 - x)(x + 1)^3
    4. The function is a quintic with a positive leading coefficient so \( y \rightarrow -\infty \) as \( x \rightarrow -\infty \) and \( y \rightarrow \infty \) as \( x \rightarrow \infty \). There is a zero at \( x = 0 \) of multiplicity \( 2 \) so a turning point occurs at \( x = 0 \). The zero at \( x = 4 \) has multiplicity \( 3 \) so a point of inflection will occur at \( x = 4 \). Thus, a possible graph is:
      Sketch of y = 2x^2(x - 4)^3
    5. The function is a quartic with a negative leading coefficient of \( -2 \) ( since \( f(x) = -2x\left(x + \frac{3}{2} \right)\left( x - 2 \right)^2 \) ), so \( y \rightarrow -\infty \) as \( x \rightarrow \pm \infty \). There are zeros of order \( 1 \) at \( x = -\frac{3}{2}, 0 \), and a zero of order \( 2 \) at \( x = 2 \). The graph will pass directly through the \( x \)-axis at \( x = -\frac{3}{2} \) and \( x = 0 \) and a turning point will occur at \( (2, 0) \). Thus, a possible graph is:
      Sketch of y = -2x(x + 3/2)(x-2)^2
  5. There is no solution provided for this question.
    1. To find the zeros, express the function in factored form.\[\begin{align*} f(x) &= -2x(x^2 - 4) \\ &= -2x(x - 2)(x + 2) \end{align*}\] Thus, when \( f(x) = 0, x = -2, 0, \) or \( 2 \). The leading coefficient is negative so the end behaviour of the cubic function can be described by \( y \rightarrow \infty \) as \( x \rightarrow -\infty \) and \( y \rightarrow -\infty \) as \( x \rightarrow \infty \) The graph of this function is
      Sketch of f(x) = -2x^3 + 8x
    2. To find the zeros, express the function in factored form.\[\begin{align*} f(x) &= -x^2(x^2 + 5x + 6) \\ &= -x^2(x + 3)(x + 2) \end{align*}\] Thus when \( f(x) = 0, x = -3, -2, \) or \( 0 \). The leading coefficient is negative so the end behaviour of the quartic function can be described by \( y \rightarrow -\infty \) as \( x \rightarrow \pm \infty \) The graph of this function is
      Sketch of f(x) = -x^4 - 5x^3 - 6x^2
    3. To find the zeros, express the function in factored form.\[\begin{align*} f(x) &= (x^2 - 1)(x^2 - 1) \\ &= (x - 1)(x + 1)(x + 1)(x - 1) \\ &= (x - 1)^2(x + 1)^2 \end{align*}\] Thus when \( f(x) = 0, x = -1 \) or \( x=1 \). The leading coefficient is positive so the end behaviour of the quartic function can be described by \( y \rightarrow \infty \) as \( x \rightarrow \pm \infty \) The graph of this function is
      Sketch of f(x) = x^4 - 2x^2 + 1
  7. There is no solution provided for this question.
    1. The function is above the \( x \)-axis (\( y \) is positive) for \( x \in (-4, -1) \cup (2, \infty) \). The function is below the \( x \)-axis (\( y \) is negative) for \( x \in (-\infty, -4) \cup (-1, 2) \).
      Sketch of y = (x - 2)(x + 1)(x + 4)
      Thus, the function is positive for \( x \in (-4, -1) \cup (2, \infty) \) and is negative for \( x \in (-\infty, -4) \cup (-1, 2) \).
    2. The function is above the \( x \)-axis for \( x \in (0, 4) \). The function is below the \( x \)-axis for \( x \in (-\infty, -3) \cup (-3, 0) \cup (4, \infty) \).
      Sketch of y = -2x(x - 4)(x + 3)^2
      Thus, the function is positive for \( x \in (0, 4) \) and is negative for \( x \in (-\infty, -3) \cup (-3, 0) \cup (4, \infty) \).
    3. The function is above the \( x \)-axis for \( x \in (-\infty, 0) \cup (0, 3) \) and is below the \( x \)-axis for \( x \in (3, \infty) \).
      Sketch of y = -x^2(x - 3)^3
      Thus, the function is positive for \( x \in (-\infty, 0) \cup (0, 3) \) and is negative for \( x \in (3, \infty) \).
  9. There is no solution provided for this question.
    1. \( x = 1 \) and \( x = 4 \) are zeros of multiplicity \( 2 \), since turning points occurs at these \( x \)-intercepts. \( x = -3 \) must therefore, be a zero of multiplicty \( 1 \) since the function is a 5th degree polynomial function. This quintic function must be of the form \( f(x) = k(x + 3)(x - 1)^2(x - 4)^2 \).
    2. The general family is \( f(x) = k(x + 3)(x - 1)^2(x - 4)^2 \). To produce the desired end behaviour, \( k \) must be negative. Choosing \( k = -1, -2 \) gives two possible members as \( f(x) = -(x + 3)(x - 1)(x - 4)^2, g(x) = -2(x + 3)(x - 1)(x - 4)^2 \).
  11. There is no solution provided for this question.