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

The vertices f(x)=2x^2 is Minimum (0,0)

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vertices f(x)=2x^2

vertices

f (x) = 2 x^{2}

Minimum (0, 0)

Solution steps

y = 2 x^{2}

The parabola parameters are:

a = 2, b = 0, c = 0

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

x_{v} = - \frac{0}{2 \cdot 2}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (0, 0)

x^{2} + y^{2} + 8 x + 4 y - 3 = 40

(x - 1)^{2} = - 4 (y - 2)

\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1

x^{2} + \frac{y ^{2}}{49} = 1

\frac{x ^{2}}{4} - \frac{y ^{2}}{25} = 1