How do you find $h (x - 4)$ $h(x-4)$ given $h (x) = x^{2} - x$ $h(x)=x^2-x$ ?

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How do you find $h (x - 4)$ $h(x-4)$ given $h (x) = x^{2} - x$ $h(x)=x^2-x$ ?

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Answer 1 · 2016-10-16T12:08:36

$h (x - 4) = x^{2} - 9 x + 20$ $h(x-4)=x^2-9x+20$

Explanation:

$h (x) = x^{2} - x$ $h(x)=x^2-x$
For $h (x - 4)$ $h(x-4)$ every $x$ $x$ is replaced by $(x - 4)$ $(x-4)$
So $h (x - 4) = {(x - 4)}^{2} - (x - 4)$ $h(x-4)=(x-4)^2-(x-4)$
This is $x^{2} - 8 x + 16 - x + 4$ $x^2-8x+16-x+4$
Or $x^{2} - 9 x + 20$ $x^2-9x+20$

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How do you find $h (x - 4)$ $h(x-4)$ given $h (x) = x^{2} - x$ $h(x)=x^2-x$ ?

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How do you find h(x−4)h(x-4) given h(x)=x2−xh(x)=x^2-x?

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How do you find $h (x - 4)$ $h(x-4)$ given $h (x) = x^{2} - x$ $h(x)=x^2-x$ ?