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

The vertices f(x)=2x^2+5x+2 is Minimum (-5/4 ,-9/8)

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

vertices

f (x) = 2 x^{2} + 5 x + 2

Minimum (- \frac{5}{4}, - \frac{9}{8})

Solution steps

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

The parabola parameters are:

a = 2, b = 5, c = 2

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{9}{8}

Therefore the parabola vertex is

(- \frac{5}{4}, - \frac{9}{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{9}{8})

f (x) = 8 x^{2}

y = 3 x - x^{2}

y = 3 x^{2} + 12 x - 3

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

f (x) = 3 x^{2} - 5

What is the vertices f(x)=2x^2+5x+2 ?
The vertices f(x)=2x^2+5x+2 is Minimum (-5/4 ,-9/8)