What is the vertices-x^2+14x-45 ?

The vertices-x^2+14x-45 is Maximum (7,4)

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vertices-x^2+14x-45

vertices

- x^{2} + 14 x - 45

Maximum (7, 4)

Solution steps

y = - x^{2} + 14 x - 45

The parabola parameters are:

a = - 1, b = 14, c = - 45

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

x_{v} = - \frac{14}{2 ( - 1 )}

Simplify

- \frac{14}{2 ( - 1 )} : 7

x_{v} = 7

Plug in

x_{v} = 7

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(7, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (7, 4)

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

y = 3 x^{2} - x - 2

y = x^{2} + 16 x + 71

f (x) = 2 x^{2} - 12 x + 1

f (x) = x^{2} + 9 x + 8