What is the vertices f(x)=2x^2+5x-3 ?

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

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vertices f(x)=2x^2+5x-3

vertices

f (x) = 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})

4 (x + \frac{3}{2})^{2} + 15

f (x) = 2 x^{2} - 8 x

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

y^{2} = - 24 x

f (x) = 2 x^{2} - 12 x + 23

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