Answer:
[tex]f(x)=\dfrac{-50(x-1)(x-2)}{(x-5)(x+5)}[/tex]
Step-by-step explanation:
Since f(x) has asymptotes at x = 5 and x = -5, then the denominator of the rational function contains the terms (x - 5) and (x + 5):
[tex]f(x)=\dfrac{?}{(x-5)(x+5)}[/tex]
Since f(x) has x-intercepts at x = 2 and x = 1, then the numerator of the rational function contains the terms (x - 2) and (x - 1):
[tex]f(x)=\dfrac{A(x-1)(x-2)}{(x-5)(x+5)}[/tex]
Now substitute the point (0, 4) and solve for A:
[tex]f(0)=4\\\\\implies \dfrac{A(0-1)(0-2)}{(0-5)(0+5)}=4\\\\\\\implies -\dfrac{2}{25}A=4\\\\\\\implies A=-50[/tex]
So final rational function:
[tex]f(x)=\dfrac{-50(x-1)(x-2)}{(x-5)(x+5)}[/tex]
