What is the vertices y=2x^2-13x-7 ?

The vertices y=2x^2-13x-7 is Minimum (13/4 ,-225/8)

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vertices y=2x^2-13x-7

vertices

y = 2 x^{2} - 13 x - 7

Minimum (\frac{13}{4}, - \frac{225}{8})

Solution steps

y = 2 x^{2} - 13 x - 7

The parabola parameters are:

a = 2, b = - 13, c = - 7

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

x_{v} = - \frac{( - 13 )}{2 \cdot 2}

Simplify

x_{v} = \frac{13}{4}

Plug in

x_{v} = \frac{13}{4}

to find the

y_{v}

value

y_{v} = - \frac{225}{8}

Therefore the parabola vertex is

(\frac{13}{4}, - \frac{225}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (\frac{13}{4}, - \frac{225}{8})

y = (x - 1)^{2} + 1

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

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

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

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

What is the vertices y=2x^2-13x-7 ?
The vertices y=2x^2-13x-7 is Minimum (13/4 ,-225/8)