How do you find the x intercept of $f (x) = \frac{x^{2} + 9 x}{x^{2} + 3 x - 4}$ $f(x) = (x^2+9x)/(x^2+3x-4)$ ?

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How do you find the x intercept of $f (x) = \frac{x^{2} + 9 x}{x^{2} + 3 x - 4}$ $f(x) = (x^2+9x)/(x^2+3x-4)$ ?

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Answer 1 · 2015-02-03T15:52:10

The $x$ $x$ -intercept is when $y = 0$ $y=0$

So at first, you only have to look at the top half of the equation -- when this is $0$ $0$ the whole thing is $0$ $0$

Leaves us with $x^{2} + 9 x = 0$ $x^2+9x=0$

This can be re-written: $x^{2} + 9 x = x (x + 9) = 0$ $x^2+9x=x(x+9)=0$

So we have two answers:
$x = 0 or (x + 9) = 0 \to x = - 9$ $x=0 or (x+9)=0->x=-9$

Check the answers against the numerator (it may not be $0$ $0$ !)
$x = 0 \to x^{2} + 3 x - 4 = - 4$ $x=0->x^2+3x-4=-4$ which is OK
$x = - 9 \to x^{2} + 3 x - 4 = 50$ $x=-9->x^2+3x-4=50$ which is also $\neq 0$ $!=0$

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How do you find the x intercept of $f (x) = \frac{x^{2} + 9 x}{x^{2} + 3 x - 4}$ $f(x) = (x^2+9x)/(x^2+3x-4)$ ?

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How do you find the x intercept of $f (x) = \frac{x^{2} + 9 x}{x^{2} + 3 x - 4}$ $f(x) = (x^2+9x)/(x^2+3x-4)$ ?