What is the vertices f(x)=x^2+4x-5 ?

The vertices f(x)=x^2+4x-5 is Minimum (-2,-9)

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vertices f(x)=x^2+4x-5

vertices

f (x) = x^{2} + 4 x - 5

Minimum (- 2, - 9)

Solution steps

y = x^{2} + 4 x - 5

The parabola parameters are:

a = 1, b = 4, c = - 5

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} = - 9

Therefore the parabola vertex is

(- 2, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 9)

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{9} = 1

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

x^{2} + 4 y^{2} - 6 x + 20 y - 2 = 0

What is the vertices f(x)=x^2+4x-5 ?
The vertices f(x)=x^2+4x-5 is Minimum (-2,-9)