How do you find the inverse of $h (x) = x^{2} - 4 x + 5$ $h(x)= x^2 - 4x + 5$ and is it a function?

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Answer 1 · 2016-07-19T09:25:39

Inversion gives two forms for $x = 2 \pm \sqrt{h - 1}, h \geq 1$ $x = 2+-sqrt(h-1), h>=1$

The for this equation is a parabola

${(x - 2)}^{2} = y - 1 \geq 0, and$ $(x-2)^2=y-1>=0, and$ so, $y \geq 1$ $y>=1$ ,

The parabola is symmetrical about its axis x = 2.

Explicitly, the inverse functions are

$x = 2 \pm \sqrt{h - 1}, h \geq 1$ $x = 2+-sqrt(h-1), h>=1$ , representing the halves of the parabola,

bifurcated by x = 2.l

How do you find the inverse of h(x)= x^2 - 4x + 5h(x)=x2−4x+5h(x)= x^2 - 4x + 5 and is it a function?