Respuesta :
Placing the quadratic function in standard form, it is found that the vertex is: (2,-16).
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Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
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In this question, the function is:
[tex]f(x) = (x - 6)(x + 2)[/tex]
To find the vertex, the function has to be in standard form. Thus:
[tex]f(x) = x^2 - 6x + 2x - 12 = x^2 - 4x - 12[/tex]
Which has coefficients [tex]a = 1, b = -4, c = -12[/tex].
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Thus, for the vertex:
[tex]\Delta = b^2-4ac = (-4)2 - 4(1)(-12) = 64[/tex]
[tex]x_{v} = -\frac{-4}{2} = 2[/tex]
[tex]y_{v} = -\frac{64}{4} = -16[/tex]
The vertex of the quadratic function is (2,-16).
A similar problem is given at https://brainly.com/question/23484680
