What is the vertices f(x)=2x^2+18 ?

The vertices f(x)=2x^2+18 is Minimum (0,18)

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vertices f(x)=2x^2+18

vertices

f (x) = 2 x^{2} + 18

Minimum (0, 18)

Solution steps

y = 2 x^{2} + 18

The parabola parameters are:

a = 2, b = 0, c = 18

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} = 18

Therefore the parabola vertex is

(0, 18)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (0, 18)

(r - 14)^{2} - 56.25

y = 7 - 6 x^{2}

y = 2 x^{2} - 1 + 5 x - 6

f (x) = - \frac{5}{8} x^{2} + \frac{1}{4} x - 2

f (x) = - 16 x^{2} + 4 x + 10