Question

Find the equation for a polynomial f(x) that satisfies the following:Degree 3- Zero at x =- Zero at 2 = -4- Zero at x = -3y-intercept of (0,12)

Answer 1

SOLUTION

We want to get a degree 3 polynomial with zeros at

$\begin{gathered} x=-3 \\ x=-4 \\ x=-3 \\ \text{and y-intercept of (0, 12)} \end{gathered}$

This polynomial can be interpreted as

$\begin{gathered} x+3=0 \\ x+4=0 \\ x+3=0 \end{gathered}$

Writing as a function we have

$\begin{gathered} y=a(x+3)(x+4)(x+3)_{} \\ y=a(x+3)(x+3)(x+4)_{} \\ y=a(x+3)^2(x+4) \end{gathered}$

So, now we need to find the value of a from the point (0, 12), so here x = 0 and y = 12, we have

$\begin{gathered} y=a(x+3)^2(x+4) \\ 12=a(0+3)^2(0+4) \\ 12=a(3)^2(4) \\ 12=a\times9\times4 \\ 12=36a \\ a=\frac{12}{36} \\ a=\frac{1}{3} \end{gathered}$

Placing the value of a into the equation

$y=a(x+3)^2(x+4)$

We have

$undefined$