What is the vertices y=2x^2-1+5x-6 ?

The vertices y=2x^2-1+5x-6 is Minimum (-5/4 ,-81/8)

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vertices y=2x^2-1+5x-6

vertices

y = 2 x^{2} - 1 + 5 x - 6

Minimum (- \frac{5}{4}, - \frac{81}{8})

Solution steps

y = 2 x^{2} - 1 + 5 x - 6

Rewrite

y = 2 x^{2} - 1 + 5 x - 6

in the form

y = a x^{2} + b x + c

y = 2 x^{2} + 5 x - 7

The parabola parameters are:

a = 2, b = 5, c = - 7

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

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

Simplify

x_{v} = - \frac{5}{4}

Plug in

x_{v} = - \frac{5}{4}

to find the

y_{v}

value

y_{v} = - \frac{81}{8}

Therefore the parabola vertex is

(- \frac{5}{4}, - \frac{81}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{5}{4}, - \frac{81}{8})

f (x) = - \frac{5}{8} x^{2} + \frac{1}{4} x - 2

f (x) = - 16 x^{2} + 4 x + 10

y = (x + 1)^{2}

- x^{2} + 4 x + 5

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

What is the vertices y=2x^2-1+5x-6 ?
The vertices y=2x^2-1+5x-6 is Minimum (-5/4 ,-81/8)