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kakasveta [241]
3 years ago
10

What are the vertical asymptotes of the function f(x) = 4x+8/x^2+3x-4

Mathematics
2 answers:
FinnZ [79.3K]3 years ago
8 0

Answer:

vertical asymptote at x=-4, 1

Step-by-step explanation:

f(x) =\frac{4x+8}{x^2+3x-4}

To find vertical asymptote we set the denominator =0 and solve for x

x^2+3x-4=0

Product is -4 and sum is +3

WE find out two factors whose sum is +3 and product is -4

4 \ times \ (-1)=-4

4-1= 3

x^2+3x-4=0

(x+4)(x-1)=0

Set each factor =0 and solve for x

x+4=0, x=-4

x-1=0, x=1

vertical asymptote at x=-4, 1

Gemiola [76]3 years ago
6 0
Vertical asymptotes occur  when the denominator = 0.

f(x) = 4x + 8 / (x + 4)(x - 1)

when x = -4 and x = 1 the denominator = 0

so the required asymptotes are x = -4 and x = 1
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