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jacob193

Answer:

The vertex of [tex]y = 4\, (x - 3)^{2} - 8[/tex] is at [tex](3,\, -8)[/tex].

Step-by-step explanation:

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

For example, the quadratic function [tex]y = 4\, (x - 3)^{2} - 8[/tex] in this question is expressed in this vertex form:

[tex]y = 4\, (x - 3)^{2} + (-8)[/tex].

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

BasketballEinstein

Answer:

[tex]\displaystyle [3, -8][/tex]

Explanation:

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

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