What is the vertices y=4x^2+5x-7 ?

The vertices y=4x^2+5x-7 is Minimum (-5/8 ,-137/16)

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vertices y=4x^2+5x-7

vertices

y = 4 x^{2} + 5 x - 7

Minimum (- \frac{5}{8}, - \frac{137}{16})

Solution steps

y = 4 x^{2} + 5 x - 7

The parabola parameters are:

a = 4, b = 5, c = - 7

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

x_{v} = - \frac{5}{2 \cdot 4}

Simplify

x_{v} = - \frac{5}{8}

Plug in

x_{v} = - \frac{5}{8}

to find the

y_{v}

value

y_{v} = - \frac{137}{16}

Therefore the parabola vertex is

(- \frac{5}{8}, - \frac{137}{16})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- \frac{5}{8}, - \frac{137}{16})

y = x^{2} + x + 8

x = - \frac{1}{4} (y + 2)^{2}

2 x^{2} - 4 x + 7

x^{2} - 5 x - 84

x^{2} + 4 x - 24 y - 92 = 0

What is the vertices y=4x^2+5x-7 ?
The vertices y=4x^2+5x-7 is Minimum (-5/8 ,-137/16)