We have the x-intercepts of the function x = -6 & x = 3, and the function passes y (0, -9).
We use the expression:
[tex]f(x)=a(x-r_1)(x-r_2)[/tex]Here a is a number, and r1 & r2 will be our x intercepts. So we replace them in the function:
[tex]f(x)=a(x+4)(x-3)[/tex]Since we have the y intercept given to us by the point (0, -9), we do as follows:
[tex]f(0)=a(4)(-3)=-9\Rightarrow a=\frac{3}{4}[/tex]So, we replace the values in our function:
[tex]f(x)=\frac{3}{4}(x+4)(x-3)\Rightarrow f(x)=\frac{3x^2}{4}+\frac{3x}{4}-9[/tex]So, that's our quadratic function that passes by (0, -9) and has x-intercepts at x = -6 & x = 3.
[tex]y=\frac{3x^2}{4}+\frac{3x}{4}-9[/tex]