Answer
$p(x)=x^3-3x^2+4$
Work Step by Step
The graph intercepts the x-axis at $x=-1$ and is tangent to the x-axis at $x=2$.
It means that the polynomial has a zero of $x=-1$ with odd multiplicity and zero of $x=2$ with even multiplicity.
Since the polynomial has a degree of 3, the multiplicity of $-1$ and $2$ is $1$ and $2$, respectively.
Then,
$p(x)=A(x+1)(x-2)^2$
$p(x)=A(x+1)(x^2-4x+4)$
$p(x)=A(x^3-3x^2+4)$
$p(x)=Ax^3-3Ax^2+4A$
The graph intercepts the y-axis at $y=4$, it means that the constant term is 4.
So, $4A=4\to A=1$.
Therefore, $p(x)=x^3-3x^2+4$.
