What is the vertices f(x)=4x^2-32x-7 ?

The vertices f(x)=4x^2-32x-7 is Minimum (4,-71)

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vertices f(x)=4x^2-32x-7

vertices

f (x) = 4 x^{2} - 32 x - 7

Minimum (4, - 71)

Solution steps

y = 4 x^{2} - 32 x - 7

The parabola parameters are:

a = 4, b = - 32, c = - 7

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

x_{v} = - \frac{( - 32 )}{2 \cdot 4}

Simplify

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 71

Therefore the parabola vertex is

(4, - 71)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (4, - 71)

2 (x + 1)^{2}

t^{2} + 12 t - 18

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

10 y^{2} = 22 x

2 x^{2} + 5 x - 7

What is the vertices f(x)=4x^2-32x-7 ?
The vertices f(x)=4x^2-32x-7 is Minimum (4,-71)