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

The vertices f(x)=x^2+5x-2 is Minimum (-5/2 ,-33/4)

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

vertices

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

Minimum (- \frac{5}{2}, - \frac{33}{4})

Solution steps

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

The parabola parameters are:

a = 1, b = 5, c = - 2

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

x_{v} = - \frac{5}{2 \cdot 1}

Simplify

x_{v} = - \frac{5}{2}

Plug in

x_{v} = - \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{33}{4}

Therefore the parabola vertex is

(- \frac{5}{2}, - \frac{33}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{5}{2}, - \frac{33}{4})

x^{2} + 6 x + 3 y + 2 = 0

y = - x^{2} + 8 x - 7

- 3 x^{2} + 6 x + 5

f (x) = - x^{2} + 12 x - 35

f (x) = 5 x^{2} - 2 x - 1

What is the vertices f(x)=x^2+5x-2 ?
The vertices f(x)=x^2+5x-2 is Minimum (-5/2 ,-33/4)