Question

write a function in any form that would match the graph shown below.
answer
attempt 1 out of 2
f(x) =
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Explanation:

Step1: Identify x-intercepts

The graph crosses the x-axis at $x=-3$, $x=0$, and $x=1$. So the factors of the polynomial are $(x+3)$, $x$, and $(x-1)$. Since the graph touches and turns at $x=-3$ and $x=1$, these are double roots. The initial form is:
$f(x)=ax^2(x+3)^2(x-1)^2$

Step2: Use y-intercept to find a

The graph passes through $(0,-3)$. Substitute $x=0$, $f(0)=-3$:
$-3=a(0)^2(0+3)^2(0-1)^2$
This is invalid, so use the point $(0,-3)$ on the curve, correct form: the graph is a quartic (degree 4, ends down), roots at $x=-3$ (double), $x=1$ (double). So $f(x)=a(x+3)^2(x-1)^2$. Substitute $(0,-3)$:
$-3=a(0+3)^2(0-1)^2$
$-3=a(9)(1)$
$a=-\frac{1}{3}$

Step3: Expand or keep factored form

Substitute $a$ into the function:
$f(x)=-\frac{1}{3}(x+3)^2(x-1)^2$
We can expand to standard form (optional):
First expand $(x+3)^2=x^2+6x+9$, $(x-1)^2=x^2-2x+1$
Multiply these: $(x^2+6x+9)(x^2-2x+1)=x^4+4x^3-2x^2-12x+9$
Then multiply by $-\frac{1}{3}$:
$f(x)=-\frac{1}{3}x^4-\frac{4}{3}x^3+\frac{2}{3}x^2+4x-3$

Answer:

$f(x)=-\frac{1}{3}(x+3)^2(x-1)^2$ or $f(x)=-\frac{1}{3}x^4-\frac{4}{3}x^3+\frac{2}{3}x^2+4x-3$