example 2\nfind another quadratic (quadratic family) with zeros -4 and 3 that passes through (-1,24).

Explanation:

Step1: Write the general form of quadratic

A quadratic function with zeros $x_1$ and $x_2$ can be written as $y = a(x - x_1)(x - x_2)$. Here $x_1=-4$ and $x_2 = 3$, so $y=a(x + 4)(x - 3)$.

Step2: Substitute the given point

Substitute the point $(-1,24)$ into the equation. When $x=-1$ and $y = 24$, we have $24=a(-1 + 4)(-1-3)$.

Step3: Solve for $a$

First, simplify the right - hand side: $(-1 + 4)(-1-3)=3\times(-4)=-12$. So the equation becomes $24=a\times(-12)$. Solving for $a$ gives $a=-2$.

Step4: Write the quadratic function

Substitute $a = - 2$ back into $y=a(x + 4)(x - 3)$. We get $y=-2(x + 4)(x - 3)=-2(x^{2}+x - 12)=-2x^{2}-2x + 24$.

Answer:

$y=-2x^{2}-2x + 24$