y = x^2 - 10x + 25 - 25
y = (x-5)^2 - 25
y+25 = (x-5)^2
x-5 = +/-sqrt(y+25)
And you get TWO inverses:
x = 5 + sqrt(y+25), for x>=5
x = 5 - sqrt(y+25), for x<=5
Answer:
[tex]y=5\pm \sqrt{(25+x)}[/tex]
Step-by-step explanation:
We are asked to find the inverse for the function [tex]y=x^2-10x[/tex].
We know that to find inverse, we interchange x and y values and then solve for y.
After interchanging x and y values, we will get:
[tex]x=y^2-10y[/tex]
Switch sides:
[tex]y^2-10y=x[/tex]
[tex]y^2-10y-x=x-x[/tex]
[tex]y^2-10y-x=0[/tex]
Now, we will use quadratic formula to solve for y.
[tex]y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]y=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(-x)}}{2(1)}[/tex]
[tex]y=\frac{10\pm \sqrt{100+4x}}{2}[/tex]
[tex]y=\frac{10\pm \sqrt{4*25+4x}}{2}[/tex]
[tex]y=\frac{10\pm \sqrt{4(25+x)}}{2}[/tex]
[tex]y=\frac{10\pm 2\sqrt{(25+x)}}{2}[/tex]
[tex]y=5\pm \sqrt{(25+x)}[/tex]
Therefore, the inverse function for our given function would be [tex]y=5\pm \sqrt{(25+x)}[/tex].
