Question

the image text is unclear, but appears to involve a math problem with points like (-1,0), (1,0), (0,-1) and a square box, likely a graphing or equation problem.

Explanation:

Step1: Use factored form of quadratic

A quadratic with x-intercepts $x=r_1$ and $x=r_2$ has form $y=a(x-r_1)(x-r_2)$. Substitute $r_1=-1$, $r_2=3$:
$y=a(x+1)(x-3)$

Step2: Solve for coefficient $a$

Substitute y-intercept $(0,-4)$ into the equation:
$-4=a(0+1)(0-3)$
$-4=a(1)(-3)$
$-4=-3a$
$a=\frac{4}{3}$

Step3: Expand to standard form

Substitute $a=\frac{4}{3}$ back and expand:
$y=\frac{4}{3}(x+1)(x-3)$
First multiply $(x+1)(x-3)=x^2-2x-3$, then distribute:
$y=\frac{4}{3}x^2 - \frac{8}{3}x - 4$

Answer:

$y=\frac{4}{3}x^2 - \frac{8}{3}x - 4$