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