What is the vertices f(x)=6x^2-5 ?

The vertices f(x)=6x^2-5 is Minimum (0,-5)

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vertices f(x)=6x^2-5

vertices

f (x) = 6 x^{2} - 5

Minimum (0, - 5)

Solution steps

y = 6 x^{2} - 5

The parabola parameters are:

a = 6, b = 0, c = - 5

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

x_{v} = - \frac{0}{2 \cdot 6}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 5

Therefore the parabola vertex is

(0, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 6

Minimum (0, - 5)

f (x) = - 2 x^{2} + 4 x + 1

3 x^{2} + 19 x - 40

(x + 4)^{2} = 12 y - 24

x^{2} - 10 x - 4 y + 21 = 0

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