What is the vertices f(x)=10x^2-17 ?

The vertices f(x)=10x^2-17 is Minimum (0,-17)

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vertices f(x)=10x^2-17

vertices

f (x) = 10 x^{2} - 17

Minimum (0, - 17)

Solution steps

y = 10 x^{2} - 17

The parabola parameters are:

a = 10, b = 0, c = - 17

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

x_{v} = - \frac{0}{2 \cdot 10}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 17

Therefore the parabola vertex is

(0, - 17)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 10

Minimum (0, - 17)

3 x^{2} - 5 x + 2

y = - 3 x^{2} + 12 x - 8

x^{2} + 8 x + 18

x = - 2 (y + 3)^{2} + 4

x^{2} + 5 x - 14

What is the vertices f(x)=10x^2-17 ?
The vertices f(x)=10x^2-17 is Minimum (0,-17)