If f(x) = x2-1, what is the equation for f'(x)? f'(x)\x+1
f-1(x) = ± √x-1
f-'(x) = ++1
4-1(x) = ± √x-1

only 10/60 isn't equivalent

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h is the x-coordinate of the vertex

step-by-step explanation:

we have

in this function

the point represent the vertex of the function

so

h is the x-coordinate of the vertex

k is the y-coordinate of the vertex