What is the vertices f(x)=2x^2+7x+3 ?

The vertices f(x)=2x^2+7x+3 is Minimum (-7/4 ,-25/8)

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vertices f(x)=2x^2+7x+3

vertices

f (x) = 2 x^{2} + 7 x + 3

Minimum (- \frac{7}{4}, - \frac{25}{8})

Solution steps

y = 2 x^{2} + 7 x + 3

The parabola parameters are:

a = 2, b = 7, c = 3

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

x_{v} = - \frac{7}{2 \cdot 2}

Simplify

x_{v} = - \frac{7}{4}

Plug in

x_{v} = - \frac{7}{4}

to find the

y_{v}

value

y_{v} = - \frac{25}{8}

Therefore the parabola vertex is

(- \frac{7}{4}, - \frac{25}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{7}{4}, - \frac{25}{8})

x = - y^{2}

f (x) = - x^{2} - 8 x - 7

4 x^{2} - 3 x + 5

f (x) = 5 x + 3 x^{2} - 4

f (x) = x^{2} + 40 x + 20

What is the vertices f(x)=2x^2+7x+3 ?
The vertices f(x)=2x^2+7x+3 is Minimum (-7/4 ,-25/8)