The quadratic equation with the given conditions is:
[tex]f(x) = \frac{3}{16}(x^2 + 2x - 24)[/tex]
The standard form of a quadratic equation with roots [tex](x_1,0)[/tex] and [tex](x_2,0)[/tex] is given by:
[tex]f(x) = a(x - x_1)(x - x_2)[/tex]
In which a is the leading coefficient.
In this problem, the roots are [tex]x_1 = -6, x_2 = 4[/tex], thus:
[tex]f(x) = a(x - (-6))(x - 4)[/tex]
[tex]f(x) = a(x + 6)(x - 4)[/tex]
[tex]f(x) = a(x^2 + 2x - 24)[/tex]
It passes through point (2,-3), which means that when [tex]x = 2, f(x) = -3[/tex], and this is used to find a.
[tex]a(2^2 + (2)(2) - 24) = -3[/tex]
[tex]-16a = -3[/tex]
[tex]a = \frac{3}{16}[/tex]
Thus, the equation is:
[tex]f(x) = \frac{3}{16}(x^2 + 2x - 24)[/tex]
A similar problem is given at https://brainly.com/question/17987697
