What is the vertices f(x)=2x^2+3x+1 ?

The vertices f(x)=2x^2+3x+1 is Minimum (-3/4 ,-1/8)

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vertices f(x)=2x^2+3x+1

vertices

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

Minimum (- \frac{3}{4}, - \frac{1}{8})

Solution steps

y = 2 x^{2} + 3 x + 1

The parabola parameters are:

a = 2, b = 3, c = 1

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

x_{v} = - \frac{3}{2 \cdot 2}

Simplify

x_{v} = - \frac{3}{4}

Plug in

x_{v} = - \frac{3}{4}

to find the

y_{v}

value

y_{v} = - \frac{1}{8}

Therefore the parabola vertex is

(- \frac{3}{4}, - \frac{1}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{3}{4}, - \frac{1}{8})

- (x + 2)^{2} + 6

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

y = - 4 (x - 2)^{2} + 4

4 x - y^{2} + y - \frac{31}{4} = 0

f (x) = x^{2} - 10 x + 16

What is the vertices f(x)=2x^2+3x+1 ?
The vertices f(x)=2x^2+3x+1 is Minimum (-3/4 ,-1/8)