What is the vertices y=-6(x-4)^2-1 ?

The vertices y=-6(x-4)^2-1 is Maximum (4,-1)

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vertices y=-6(x-4)^2-1

vertices

y = - 6 (x - 4)^{2} - 1

Maximum (4, - 1)

Solution steps

y = - 6 (x - 4)^{2} - 1

Rewrite

y = - 6 (x - 4)^{2} - 1

in the form

y = a x^{2} + b x + c

y = - 6 x^{2} + 48 x - 97

The parabola parameters are:

a = - 6, b = 48, c = - 97

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

x_{v} = - \frac{48}{2 ( - 6 )}

Simplify

- \frac{48}{2 ( - 6 )} : 4

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(4, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 6

Maximum (4, - 1)

f (x) = \frac{x ^{2}}{x + 2}

f (x) = (\frac{2}{x}) (\frac{x}{x + 2})

4 x - 3

8 y - 4 x = - 56

x^{3} + 10 x

What is the vertices y=-6(x-4)^2-1 ?
The vertices y=-6(x-4)^2-1 is Maximum (4,-1)