What are the vertical asymptotes of the function f(x) = the quantity of 4 x plus 8, all over x squared plus 3 x minus 4?

1 answer:

5 0

4x+8/x^2+3x-4
Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function.

the vertical asymptotes (which tell me where the graph can not go) will be at the values
4x+8/x^2+3x-4

if u solve the denominator x should not eqaul to 0

so x^2+3x-4=0 x=1,x=-4

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Step-by-step explanation:

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The given Polynomial is :

$f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})$

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$\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4$

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So, no root of this polynomial is real.

Therefore, All the four roots of Polynomial are imaginary.

So, we can't say whether the number k=2, is an upper or lower bound of the polynomial $f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})$ .