What is the inverse of y=(x^2-6x-7)/(x+5) ?

The inverse of y=(x^2-6x-7)/(x+5) is (6+x+sqrt(x^2+32x+64))/2 ,(6+x-sqrt(x^2+32x+64))/2

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inverse of y=(x^2-6x-7)/(x+5)

inverse

y = \frac{x ^{2} - 6 x - 7}{x + 5}

\frac{6 + x + x ^{2} + 32 x + 64}{2}, \frac{6 + x - x ^{2} + 32 x + 64}{2}

Solution steps

y = \frac{x ^{2} - 6 x - 7}{x + 5}

Replace

x

with

y

x = \frac{y ^{2} - 6 y - 7}{y + 5}

Solve for

y, x = \frac{y ^{2} - 6 y - 7}{y + 5}

\frac{6 + x + x ^{2} + 32 x + 64}{2}, \frac{6 + x - x ^{2} + 32 x + 64}{2}

\frac{( 9 x - 4 )}{4 x + 1}

\frac{1}{x ^{2} + 4 x + 4}

x^{23} - 3

f (x) = \frac{7 x + 9}{6 x - 7}

f (x) = 8^{2} - 15 x + 8

What is the inverse of y=(x^2-6x-7)/(x+5) ?
The inverse of y=(x^2-6x-7)/(x+5) is (6+x+sqrt(x^2+32x+64))/2 ,(6+x-sqrt(x^2+32x+64))/2