Question

A) Sketch a polynomial function that has the following properties:
• x-intercepts at (3, 0), (- 2, 0), (1/3, 0)
• A degree of 4
• As x →-∞, y →-∞ and as x → ∞, Y →-∞

B)Write a possible equation for the polynomial in standard form.

Answer 1

Answer:

$\textsf{A)}\quad \sf See \; attached.$

$\textsf{B)} \quad y=-3x^4+13x^3+5x^2-57x+18$

Step-by-step explanation:

Part A

A polynomial of degree n can have up to (n - 1) turning points. Therefore, given that the polynomial in question is of degree 4, there will be at most 3 turning points.

If a polynomial has 3 x-intercepts and a maximum of 3 turning points, one of the x-intercepts will have a multiplicity of two, causing the curve to bounce off the x-axis at this point rather than passing through it.

To sketch a polynomial function with the given properties:

• Plot the x-intercepts at (-2, 0), (1/3, 0) and (3, 0).
• Start the curve in quadrant III and pass it through the x-intercept (-2, 0) into quadrant II.
• Change the direction of the curve and pass it through x-intercept (1/3, 0) into quadrant IV.
• Change the direction of the curve and bounce it off the x-intercept (3, 0) so that as x → -∞, y → -∞.

Part B

Since the degree of the polynomial is 4, the factored form of the polynomial will be:

$y=a(x-r_1)(x-r_2)(x-r_3)(x-r_4)$

where a is the leading coefficient, and rₙ are the x-intercepts.

Given that the x-intercepts are x = 1/3, x = -2 and x = 3, and we have chosen to assign a multiplicity of two to the x-intercept at x = 3, then the polynomial expression is represented as:

$y=a(3x-1)(x+2)(x-3)(x-3)$

$y=a(3x-1)(x+2)(x-3)^2$

The end behaviour of a polynomial of degree 4 is:

Positive leading coefficient:

• $\textsf{As}\;x \rightarrow -\infty, f(x) \rigtharrow +\infty$
• $\textsf{As}\;x \rightarrow +\infty, f(x) \rigtharrow +\infty$

Negative leading coefficient:

• $\textsf{As}\;x \rightarrow -\infty, f(x) \rigtharrow -\infty$
• $\textsf{As}\;x \rightarrow +\infty, f(x) \rigtharrow -\infty$

Given that as x approaches negative and positive infinity, the function approaches negative infinity, we can conclude that the leading coefficient is negative, so:

$y=-(3x-1)(x+2)(x-3)^2$

To write this equation in standard form, simply expand the brackets:

$y=-(3x-1)(x+2)(x-3)(x-3)$

$y=-(3x^2+5x-2)(x^2-6x+9)$

$y=-(3x^4-18x^3+27x^2+5x^3-30x^2+45x-2x^2+12x-18)$

$y=-(3x^4-13x^3-5x^2+57x-18)$

$y=-3x^4+13x^3+5x^2-57x+18$