What is the vertices x^2-12x+16y+68=0 ?

The vertices x^2-12x+16y+68=0 is Maximum (6,-2)

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vertices x^2-12x+16y+68=0

vertices

x^{2} - 12 x + 16 y + 68 = 0

Maximum (6, - 2)

Solution steps

x^{2} - 12 x + 16 y + 68 = 0

Rewrite

x^{2} - 12 x + 16 y + 68 = 0

in the form

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

y = - \frac{x ^{2}}{16} + \frac{3 x}{4} - \frac{17}{4}

The parabola parameters are:

a = - \frac{1}{16}, b = \frac{3}{4}, c = - \frac{17}{4}

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

x_{v} = - \frac{( \frac{3}{4} )}{2 ( - \frac{1}{16} )}

Simplify

- \frac{\frac{3}{4}}{2 ( - \frac{1}{16} )} : 6

x_{v} = 6

Plug in

x_{v} = 6

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(6, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{16}

Maximum (6, - 2)

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

f (x) = x^{2} - 4 x + 7

y = x^{2} - 6 x + 10

y = 3 x^{2} - 30 x + 77

y = x^{2} - 4

What is the vertices x^2-12x+16y+68=0 ?
The vertices x^2-12x+16y+68=0 is Maximum (6,-2)