What is the vertices y=x^2-5x-14 ?

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

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vertices y=x^2-5x-14

vertices

y = 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})

x^{2} + 14 x - 8 y + 65 = 0

x^{2} - 4 x - 4

f (x) = 2 x^{2} - 6 x + 15

f (x) = 2 x^{2} + 5 x - 3

4 (x + \frac{3}{2})^{2} + 15

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