What is the vertices f(x)=-x^2+4x+7 ?

The vertices f(x)=-x^2+4x+7 is Maximum (2,11)

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vertices f(x)=-x^2+4x+7

vertices

f (x) = - x^{2} + 4 x + 7

Maximum (2, 11)

Solution steps

y = - x^{2} + 4 x + 7

The parabola parameters are:

a = - 1, b = 4, c = 7

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 ( - 1 )}

Simplify

- \frac{4}{2 ( - 1 )} : 2

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 11

Therefore the parabola vertex is

(2, 11)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 11)

f (x) = - x^{2} + 4 x - 6

f (x) = x^{2} - 12 x + 36

y = x^{2} + 4 x + 1

f (x) = x^{2} + 16 x + 63

y = 2 x^{2} - 4 x + 7

What is the vertices f(x)=-x^2+4x+7 ?
The vertices f(x)=-x^2+4x+7 is Maximum (2,11)