What is the vertices y=2x^2+5x-12 ?

The vertices y=2x^2+5x-12 is Minimum (-5/4 ,-121/8)

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vertices y=2x^2+5x-12

vertices

y = 2 x^{2} + 5 x - 12

Minimum (- \frac{5}{4}, - \frac{121}{8})

Solution steps

y = 2 x^{2} + 5 x - 12

The parabola parameters are:

a = 2, b = 5, c = - 12

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{121}{8}

Therefore the parabola vertex is

(- \frac{5}{4}, - \frac{121}{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{121}{8})

f (x) = 3 (x + 2)^{2} + 1

- 2 x^{2} + 8 x - 1

y = x^{2} + 3 x

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

x = - 3 y^{2} + 5

What is the vertices y=2x^2+5x-12 ?
The vertices y=2x^2+5x-12 is Minimum (-5/4 ,-121/8)