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# ​ Domain and range

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Please explain each step thank you

Solved: y^2-8x +7

(y-4)^2 -9

sqrt(x-9)+4 = y

Sep 9, 2019
edited by Guest  Sep 9, 2019

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g(x) is a concave up parabola

$$g(x)=x^2-8x+7\\ g(x)=(x-7)(x-1)\\ \text{roots are at x= 7 and x= 1}\\ \text{Axis of symmetry = 4}$$

x=4    g(4)= -3*3 = -9

So the vertex is (4,-9)

Since the domain is $$(4,\infty)$$     the range is  $$(-9,\infty)$$

So the domain and range of the inverse must be the other way around.

For $$g^{-1}(x)$$

The domain is $$(-9,\infty)$$  abd the range is $$(4,\infty)$$

Here is the graph.

https://www.desmos.com/calculator/lgob2ihohq

This maths underneath is not needed to answer the question but I have done it so I might as well leave it.

$$Let\;\;y=g(x)\\ y=x^2-8x+7 \qquad x>4\\ y-7=x^2-8x\\ y-7+16=x^2-8x+16\\ y+9=(x-4)^2\\ \pm\sqrt{y+9}=x-4\\ \text{But x-4 is positive, so}\\ x-4=\sqrt{y+9}\\ x=\sqrt{y+9}+4\\ g^{-1}(x)=\sqrt{x+9}+4$$

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Sep 9, 2019