What is the vertices f(x)=x^2+3x+1/4 ?

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

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vertices f(x)=x^2+3x+1/4

vertices

f (x) = x^{2} + 3 x + \frac{1}{4}

Minimum (- \frac{3}{2}, - 2)

Solution steps

y = x^{2} + 3 x + \frac{1}{4}

The parabola parameters are:

a = 1, b = 3, c = \frac{1}{4}

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

x_{v} = - \frac{3}{2 \cdot 1}

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(- \frac{3}{2}, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{3}{2}, - 2)

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

f (x) = x^{2} + 6 x + 2

f (x) = x^{2} + 6 x + 3

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

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

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