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Popular >

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

Solution

vertices

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

Solution

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

Solution steps

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

The parabola parameters are:

a = 3, b = 2, c = - 1

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

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

Simplify

- \frac{2}{2 \cdot 3} : - \frac{1}{3}

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

Plug in

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

to find the

y_{v}

value

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

Therefore the parabola vertex is

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

If

a < 0,

then the vertex is a maximum value

If

a > 0,

then the vertex is a minimum value

a = 3

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

Popular Examples

vertices x^2-5y=0 vertices

x^{2} - 5 y = 0

vertices y=2x^2-8x vertices

y = 2 x^{2} - 8 x

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

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

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

Frequently Asked Questions (FAQ)

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