What is the vertices y=-x^2+12x-35 ?

The vertices y=-x^2+12x-35 is Maximum (6,1)

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vertices y=-x^2+12x-35

vertices

y = - x^{2} + 12 x - 35

Maximum (6, 1)

Solution steps

y = - x^{2} + 12 x - 35

The parabola parameters are:

a = - 1, b = 12, c = - 35

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

x_{v} = - \frac{12}{2 ( - 1 )}

Simplify

- \frac{12}{2 ( - 1 )} : 6

x_{v} = 6

Plug in

x_{v} = 6

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(6, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (6, 1)

f (x) = - 3 sin (2 x - 10) - 3

f (x) = \frac{x - 3}{x + 7}

y = x + 1

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

y = e^{- x^{2}}

What is the vertices y=-x^2+12x-35 ?
The vertices y=-x^2+12x-35 is Maximum (6,1)