Determine an equation of the inverse of y=(x-4)^2+5?

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Determine an equation of the inverse of y=(x-4)^2+5?

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Answer 1 · 2018-08-03T10:10:09

$y = 4 \pm \sqrt{x - 5}$ $y=4+-sqrt(x-5)$

Explanation:

An inverse function can be found by swapping the $x$ $x$ and $y$ $y$

ie
$y = {(x - 4)}^{2} + 5$ $y=(x-4)^2+5$ is your original function

Swapping the $x$ $x$ and $y$ $y$ around, you'll get

$x = {(y - 4)}^{2} + 5$ $x=(y-4)^2+5$

Solving

$x = {(y - 4)}^{2} + 5$ $x=(y-4)^2+5$
$x - 5 = {(y - 4)}^{2}$ $x-5=(y-4)^2$
$\pm \sqrt{x - 5} = y - 4$ $+-sqrt(x-5)=y-4$
$y = 4 \pm \sqrt{x - 5}$ $y=4+-sqrt(x-5)$

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Determine an equation of the inverse of y=(x-4)^2+5?

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