What is the vertices y=2x^2+5x-3 ?

The vertices y=2x^2+5x-3 is Minimum (-5/4 ,-49/8)

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vertices y=2x^2+5x-3

vertices

y = 2 x^{2} + 5 x - 3

Minimum (- \frac{5}{4}, - \frac{49}{8})

Solution steps

y = 2 x^{2} + 5 x - 3

The parabola parameters are:

a = 2, b = 5, c = - 3

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{49}{8}

Therefore the parabola vertex is

(- \frac{5}{4}, - \frac{49}{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{49}{8})

x^{2} - 5 x + 6

2 x^{2} + 3

y = x^{2} + 2 x - 8

- 3 x^{2} + 5 x

y = 4 x + 4 x - 2 + x^{2} - 3

What is the vertices y=2x^2+5x-3 ?
The vertices y=2x^2+5x-3 is Minimum (-5/4 ,-49/8)