What is the vertices x^2+5x-14 ?

The vertices x^2+5x-14 is Minimum (-5/2 ,-81/4)

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vertices x^2+5x-14

vertices

x^{2} + 5 x - 14

Minimum (- \frac{5}{2}, - \frac{81}{4})

Solution steps

y = x^{2} + 5 x - 14

The parabola parameters are:

a = 1, b = 5, c = - 14

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{81}{4}

Therefore the parabola vertex is

(- \frac{5}{2}, - \frac{81}{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{81}{4})

f (x) = x^{2} + 8 x + 7

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

f (x) = 2 x^{2} + 8 x + 3

y = 2 x^{2} - 7

y = - 3 x^{2} - 36 x - 104

What is the vertices x^2+5x-14 ?
The vertices x^2+5x-14 is Minimum (-5/2 ,-81/4)