Respuesta :
Answer:
[tex]f(x)=\frac{x^2-20x+20}{x^2+3x-4}[/tex]
Step-by-step explanation:
Let the fractional form of the function is
[tex]f(x)= \frac {N}{D}\cdots(i).[/tex]
As there are two vertical asymptotes at x = -4 and x = 1
so, the denominator of the function must be 0 at x=-4 as well as x=1, so, (x+4) and (x-1) is factors of the denominator, D.
D=a(x+4)(x-1)
where a is constant.
As, at x intercepts at x = 4 and x = -5, so the function is zero at x=4 and x=-5.
So, (x-4) and (x+5) are the factors of numerator, so
N=b(x-4)(x+5), where b is a constant.
From the equation (i),
[tex]f(x)=\frac {b(x-4)(x+5)}{a(x+4)(x-1)}\cdots(ii)[/tex]
As y intercept is 5, so at x=0,
f(x=0)=5
[tex]\Rightarrow \frac {b(0-4)(0+5)}{a(0+4)(0-1)}=5[/tex]
[tex]\Rightarrow \frac b a \times \frac {-20}{-4}=5[/tex]
[tex]\Rightarrow \frac b a =1[/tex]
Put the value of [tex]b/a[/tex] in equation (ii), the required rational function is
[tex]f(x)=\frac {(x-4)(x+5)}{(x+4)(x-1)}[/tex]
[tex]\Rightarrow f(x)=\frac{x^2-20x+20}{x^2+3x-4}[/tex]
