Question

show all your workfor questions 1 – 4, given the graph of a polynomial function• use the roots of the graph to write the equation• use an additional point to find the value of a, the1.

Explanation:

Step1: Identify roots from graph

The graph crosses/touches the x-axis at $x=-2$, $x=1$, $x=3$.

Step2: Write factored polynomial form

Since the graph has a local minimum at $x=-2$ (touches x-axis, even multiplicity; degree of polynomial is 4, so $x=-2$ has multiplicity 2), the form is:
$y = a(x+2)^2(x-1)(x-3)$

Step3: Use vertex point to find $a$

The lowest vertex is $(-1, -4)$. Substitute $x=-1$, $y=-4$:
$-4 = a(-1+2)^2(-1-1)(-1-3)$
$-4 = a(1)^2(-2)(-4)$
$-4 = a(8)$
$a = \frac{-4}{8} = -\frac{1}{2}$

Step4: Expand to standard form (optional, or keep factored)

Substitute $a$ back into the factored form:
$y = -\frac{1}{2}(x+2)^2(x-1)(x-3)$
To expand fully:
First expand $(x+2)^2 = x^2+4x+4$, and $(x-1)(x-3)=x^2-4x+3$
Multiply those two quadratics:
$(x^2+4x+4)(x^2-4x+3) = x^4 -4x^3+3x^2+4x^3-16x^2+12x+4x^2-16x+12$
Simplify: $x^2-4x+3$
$= x^4 -9x^2 -4x +12$
Multiply by $-\frac{1}{2}$:
$y = -\frac{1}{2}x^4 + \frac{9}{2}x^2 + 2x -6$

Answer:

Factored form: $\boldsymbol{y = -\frac{1}{2}(x+2)^2(x-1)(x-3)}$
Standard form: $\boldsymbol{y = -\frac{1}{2}x^4 + \frac{9}{2}x^2 + 2x - 6}$