Answer:
[tex]\displaystyle y=-\frac{1}{2}(x+4)(x-2)[/tex]
Step-by-step explanation:
We want to find a quadratic that passes through the points:
[tex](-4, 0), \, (2, 0), \text{ and } (6, -20)[/tex]
In intercept form.
First, note that the first two given points are the x-intercepts of our quadratic. Intercept or factored form is given by:
[tex]y=a(x-p)(x-q)[/tex]
Where p and q are the x-intercepts, and a is the leading coefficient.
So, we will substitute -4 and 2 for p and q:
[tex]y=a(x-(-4))(x-(2))[/tex]
Simplify:
[tex]y=a(x+4)(x-2)[/tex]
Next, the third point (6, -20) tells us that y = -20 when x = 6. So:
[tex](-20)=a(6+4)(6-2)[/tex]
Solve for a:
[tex]-20=10(4)a\Rightarrow 40a=-20[/tex]
Thus:
[tex]\displaystyle a=\frac{-20}{40}=-\frac{1}{2}[/tex]
Hence, our equation in intercept from is:
[tex]\displaystyle y=-\frac{1}{2}(x+4)(x-2)[/tex]