What is the vertices f(x)=8x^2+4x-1 ?

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

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vertices f(x)=8x^2+4x-1

vertices

f (x) = 8 x^{2} + 4 x - 1

Minimum (- \frac{1}{4}, - \frac{3}{2})

Solution steps

y = 8 x^{2} + 4 x - 1

The parabola parameters are:

a = 8, b = 4, c = - 1

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

x_{v} = - \frac{4}{2 \cdot 8}

Simplify

- \frac{4}{2 \cdot 8} : - \frac{1}{4}

x_{v} = - \frac{1}{4}

Plug in

x_{v} = - \frac{1}{4}

to find the

y_{v}

value

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

Therefore the parabola vertex is

(- \frac{1}{4}, - \frac{3}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 8

Minimum (- \frac{1}{4}, - \frac{3}{2})

f (x) = - x^{2} - 14 x - 48

x^{2} + 10 x - 2 y + 23 = 0

+ x^{2} - 36

f (x) = 2 (x + 14)^{2} - 6

y = 3 x^{2} - 6 x + 5

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