Given $f (x) = x + 2, g (x) = x - 3, h (x) = x + 4$ $f(x)=x+2, g(x)=x-3, h(x)=x+4$ how do you determine $y = \frac{f (x)}{h (x)} \times \frac{g (x)}{h (x)}$ $y=(f(x))/(h(x))times(g(x))/(h(x))$ ?

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Answer 1 · 2017-07-01T11:16:13

$y = \frac{x^{2} - x - 6}{x^{2} + 8 x + 16}$ $y = frac(x^(2) - x - 6)(x^(2) + 8 x + 16)$

We have: $y = \frac{f (x)}{h (x)} \times \frac{g (x)}{h (x)}$ $y = frac(f(x))(h(x)) times frac(g(x))(h(x))$

$\Rightarrow y = \frac{f (x) \times g (x)}{{(h (x))}^{2}}$ $Rightarrow y = frac(f(x) times g(x))((h(x))^(2))$

Let's substitute the expressions for $f (x)$ $f(x)$ , $g (x)$ $g(x)$ and $h (x)$ $h(x)$ into the equation:

$\Rightarrow y = \frac{(x + 2) \times (x - 3)}{{(x + 4)}^{2}}$ $Rightarrow y = frac((x + 2) times (x - 3))((x + 4)^(2))$

Then, let's expand the parentheses:

$\Rightarrow y = \frac{x^{2} + 2 x - 3 x - 6}{x^{2} + 4 x + 4 x + 16}$ $Rightarrow y = frac(x^(2) + 2 x - 3 x - 6)(x^(2) + 4 x + 4 x + 16)$

$therefore y = frac(x^(2) - x - 6)(x^(2) + 8 x + 16)$

Given f(x)=x+2, g(x)=x-3, h(x)=x+4f(x)=x+2,g(x)=x−3,h(x)=x+4f(x)=x+2, g(x)=x-3, h(x)=x+4 how do you determine y=(f(x))/(h(x))times(g(x))/(h(x))y=f(x)h(x)×g(x)h(x)y=(f(x))/(h(x))times(g(x))/(h(x))?