Question

The graph of a cubic polynomial $y=a{x}^3+b{x}^2+cx+d$ is shown. Find the coefficients $a,b,c$ and $d$.

Answer 1

This polynomial has a zero of multiplicity $1$ at $x=-2$ and a zero of multiplicity $2$ at $x=1$. Hence the polynomial may be written as

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

This polynomial has a $y$ intercept $(0,1)$. Hence

$1=a(0+2)(0-1{)}^2$

Solve for $a$ to obtain

$a=\frac{1}{2}$

The polynomial may now be written as follows

$y=\left(\frac{1}{2}\right)(x+2){(x-1)}^2$

We now identify the coefficients by comparing the polynomial

$y=\left(\frac{1}{2}\right)x^3+x^2-\left(\frac{3}{2}\right)x+1$ by the polynomial $y=a{x}^3+b{x}^2+cx+d$ we get,

$a=\frac{1}{2}$, $b=1$, $c=-\frac{3}{2}$ and $d=1$