What is the vertices f(x)=x^2+14x-72 ?

The vertices f(x)=x^2+14x-72 is Minimum (-7,-121)

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vertices f(x)=x^2+14x-72

vertices

f (x) = x^{2} + 14 x - 72

Minimum (- 7, - 121)

Solution steps

y = x^{2} + 14 x - 72

The parabola parameters are:

a = 1, b = 14, c = - 72

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

x_{v} = - \frac{14}{2 \cdot 1}

Simplify

x_{v} = - 7

Plug in

x_{v} = - 7

to find the

y_{v}

value

y_{v} = - 121

Therefore the parabola vertex is

(- 7, - 121)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 7, - 121)

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

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

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

y = x^{2} + 12 x + 36

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

What is the vertices f(x)=x^2+14x-72 ?
The vertices f(x)=x^2+14x-72 is Minimum (-7,-121)