Question

f(x)=x^2-8x-5 what is the y-coordinate of the vertex? the minimum? what is the range of f(x)?

Answer 1

at vertex f'(x)=0
so 2x-8=0
x=4
so y=(4)^2-8(4)-5 = -21

So that;s the minimum, so y is defined on [-21,infinity)

Answer 2

what is the range then?

Answer 3

given above, -21 to positive infinity

Answer 4

what would the x coordinate be?

Answer 5

is that the )?

Answer 6

0*

Answer 7

the x coordinate at y=-21 is 4, but x can be any real number

Answer 8

so the vertex is x= 4 and y=-21?

Answer 9

yeah