Question

What is the vertex of this
quadratic function?
y = 4(x-3)2 – 8

Answer 1

`\displaystyle [3, -8]`

Explanation:

First off, this equation came from the quadratic equation
`\displaystyle [y = Ax^2 + Bx + C],`in which it was
`\displaystyle y = 4x^2 - 24x + 28.`In the vertex equation
`\displaystyle [y = A(x - h)^2 + k],`the vertex is represented by
`\displaystyle [h, k],`in which −h gives you OPPOCITE TERMS OF WHAT THEY REALLY ARE, so be careful there.

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Answer 2

The vertex of
`y = 4\, (x - 3)^(2) - 8` is at
`(3,\, -8)`.

Explanation:

Consider a quadratic function with the point
`(h,\, k)` as the vertex (for some constants
`h` and
`k`.) It would then be possible to express this function in the vertex form
`y = a\, (x - h)^(2) + k` for some constant
`a` where
`a \\e 0`. (Note the minus sign in front of
`h`.)

For example, the quadratic function
`y = 4\, (x - 3)^(2) - 8` in this question is expressed in this vertex form:

`y = 4\, (x - 3)^(2) + (-8)`.

The values of
`h` and
`k` are
`h = 3` and
`k = (-8)`, respectively. Thus, the vertex of this parabola would be at the point
`(3,\, -8)`.