What is the vertices f(x)=-x^2+6x+7 ?

The vertices f(x)=-x^2+6x+7 is Maximum (3,16)

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vertices f(x)=-x^2+6x+7

vertices

f (x) = - x^{2} + 6 x + 7

Maximum (3, 16)

Solution steps

y = - x^{2} + 6 x + 7

The parabola parameters are:

a = - 1, b = 6, c = 7

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

x_{v} = - \frac{6}{2 ( - 1 )}

Simplify

- \frac{6}{2 ( - 1 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 16

Therefore the parabola vertex is

(3, 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (3, 16)

- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1

- \frac{1}{3} (x - 3) = (y + 5)^{2}

y, x^{2} - y^{2} = 0

25 x^{2} + 9 y^{2} - 100 x + 18 y - 116 = 0

What is the vertices f(x)=-x^2+6x+7 ?
The vertices f(x)=-x^2+6x+7 is Maximum (3,16)