What is the vertices y=7-6x^2 ?

The vertices y=7-6x^2 is Maximum (0,7)

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vertices y=7-6x^2

vertices

y = 7 - 6 x^{2}

Maximum (0, 7)

Solution steps

y = 7 - 6 x^{2}

Rewrite

y = 7 - 6 x^{2}

in the form

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

y = - 6 x^{2} + 7

The parabola parameters are:

a = - 6, b = 0, c = 7

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

x_{v} = - \frac{0}{2 ( - 6 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 7

Therefore the parabola vertex is

(0, 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 6

Maximum (0, 7)

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

y = (x + 1)^{2}

- x^{2} + 4 x + 5