What is the vertices f(x)=x^2+4x-21 ?

The vertices f(x)=x^2+4x-21 is Minimum (-2,-25)

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vertices f(x)=x^2+4x-21

vertices

f (x) = x^{2} + 4 x - 21

Minimum (- 2, - 25)

Solution steps

y = x^{2} + 4 x - 21

The parabola parameters are:

a = 1, b = 4, c = - 21

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

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 25

Therefore the parabola vertex is

(- 2, - 25)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 25)

5 + 4 x - x^{2}

8 - 5 x^{2}

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

f (x) = 2 x^{2} - 4 x - 3

2 x^{2} - 1 + 5 x

What is the vertices f(x)=x^2+4x-21 ?
The vertices f(x)=x^2+4x-21 is Minimum (-2,-25)