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Popular >

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

Solution

vertices

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

Solution

Maximum (\frac{1}{4}, \frac{1}{4})

Solution steps

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

The parabola parameters are:

a = - 4, b = 2, c = 0

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

x_{v} = - \frac{2}{2 ( - 4 )}

Simplify

- \frac{2}{2 ( - 4 )} : \frac{1}{4}

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

Plug in

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

to find the

y_{v}

value

y_{v} = \frac{1}{4}

Therefore the parabola vertex is

(\frac{1}{4}, \frac{1}{4})

If

a < 0,

then the vertex is a maximum value

If

a > 0,

then the vertex is a minimum value

a = - 4

Maximum (\frac{1}{4}, \frac{1}{4})

Popular Examples

vertices y=2x^2-6x vertices

y = 2 x^{2} - 6 x

vertices f(x)=-1/3 x^2-7/6 x+1 vertices

f (x) = - \frac{1}{3} x^{2} - \frac{7}{6} x + 1

vertices 4x^2+4x-8 vertices

4 x^{2} + 4 x - 8

vertices f(x)=-2x^2-16x-31 vertices

f (x) = - 2 x^{2} - 16 x - 31

vertices-x^2+2x+2 vertices

- x^{2} + 2 x + 2

Frequently Asked Questions (FAQ)

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