What is the vertices 2x^2+5x-7 ?

The vertices 2x^2+5x-7 is Minimum (-5/4 ,-81/8)

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vertices 2x^2+5x-7

vertices

2 x^{2} + 5 x - 7

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

Solution steps

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})

x^{2} + 12 y = 0

- 4.9 t^{2} + 9.8 t

- x^{2} + 4 x + 3

y = x^{2} - 4 x - 12

(x + 4)^{2}

What is the vertices 2x^2+5x-7 ?
The vertices 2x^2+5x-7 is Minimum (-5/4 ,-81/8)