# What is the the vertex and the axis of symmetry of the parabola: y = x^2 − 16x + 63?

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We need to determine the vertex and the the axis of symmetry of the parabola y = x^2 - 16x + 63

For the general equation of a parabola y = ax^2 + bx + c, the x-coordinate of the vertex is given by -b/2a.

Substituting the values for the given parabola we have 16/2 = 8

For x = 8, y = 8^2 - 16*8 + 63 = 64 - 128 + 63 = -1

The vertex of the given parabola is (8, -1)

The axis of symmetry of a parabola is given by x = -b/2a

For the given parabola it is x = 8

**The axis of symmetry of the parabola is x = 8 and the vertex is (8, -1)**

use the formula

plug in x

vertex is

the aos is always the x of the vertex so aos is `8`

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ugh my things didnt show:

`y=x^2-16x+63`

`a= 1 b=-16 c=16`

use the formula

`x= 16/(2(1))` `x=16/2` `x=8 `

plug in x

`y=8^2-16(8)+63`

`y=64-128+63`

`64-128=-64+63=-1`

`y=-1`

vertex is` (8,1)`

the aos is always the x of the vertex so aos is `8`