Answer:
One possible equation is [tex]f(x) = (x + 3)\, (x - 5)[/tex], which is equivalent to [tex]f(x) = x^{2} - 2\, x - 15[/tex].
Step-by-step explanation:
The factor theorem states that if [tex]x = x_{0}[/tex] (where [tex]x_{0}[/tex] is a constant) is a root of a function, [tex](x - x_{0})[/tex] would be a factor of that function.
The question states that [tex](-3,\, 0)[/tex] and [tex](5,\, 0)[/tex] are [tex]x[/tex]-intercepts of this function. In other words, [tex]x = -3[/tex] and [tex]x = 5[/tex] would both set the value of this quadratic function to [tex]0[/tex]. Thus, [tex]x = -3\![/tex] and [tex]x = 5\![/tex] would be two roots of this function.
By the factor theorem, [tex](x - (-3))[/tex] and [tex](x - 5)[/tex] would be two factors of this function.
Because the function in this question is quadratic, [tex](x - (-3))[/tex] and [tex](x - 5)[/tex] would be the only two factors of this function. In other words, for some constant [tex]a[/tex] ([tex]a \ne 0[/tex]):
[tex]f(x) = a\, (x - (-3))\, (x - 5)[/tex].
Simplify to obtain:
[tex]f(x) = a\, (x + 3)\, (x - 5)[/tex].
Expand this expression to obtain:
[tex]f(x) = a\, (x^{2} - 2\, x - 15)[/tex].
(Quadratic functions are polynomials of degree two. If this function has any factor other than [tex](x - (-3))[/tex] and [tex](x - 5)[/tex], expanding the expression would give a polynomial of degree at least three- not quadratic.)
Every non-zero value of [tex]a[/tex] corresponds to a distinct quadratic function with [tex]x[/tex]-intercepts [tex](-3,\, 0)[/tex] and [tex](5,\, 0)[/tex]. For example, with [tex]a = 1[/tex]:
[tex]f(x) = (x + 3)\, (x - 5)[/tex], or equivalently,
[tex]f(x) = x^{2} - 2\, x - 15[/tex].
