What is the vertices 2x^2-1+5x ?

The vertices 2x^2-1+5x is Minimum (-5/4 ,-33/8)

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vertices 2x^2-1+5x

vertices

2 x^{2} - 1 + 5 x

Minimum (- \frac{5}{4}, - \frac{33}{8})

Solution steps

y = 2 x^{2} - 1 + 5 x

Rewrite

y = 2 x^{2} - 1 + 5 x

in the form

y = a x^{2} + b x + c

y = 2 x^{2} + 5 x - 1

The parabola parameters are:

a = 2, b = 5, c = - 1

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{33}{8}

Therefore the parabola vertex is

(- \frac{5}{4}, - \frac{33}{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{33}{8})

y = - x^{2} + 18 x - 82

+ 2 (x + 1)^{2} - 3

f (x) = x^{2} - 4 x - 21

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

y = 3 (x - 3)^{2} - 1

What is the vertices 2x^2-1+5x ?
The vertices 2x^2-1+5x is Minimum (-5/4 ,-33/8)