What is the vertices x^2+6x-8y+1=0 ?

The vertices x^2+6x-8y+1=0 is Minimum (-3,-1)

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vertices x^2+6x-8y+1=0

vertices

x^{2} + 6 x - 8 y + 1 = 0

Minimum (- 3, - 1)

Solution steps

x^{2} + 6 x - 8 y + 1 = 0

Rewrite

x^{2} + 6 x - 8 y + 1 = 0

in the form

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

y = \frac{1 \cdot x ^{2}}{8} + \frac{3 x}{4} + \frac{1}{8}

The parabola parameters are:

a = \frac{1}{8}, b = \frac{3}{4}, c = \frac{1}{8}

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

x_{v} = - \frac{( \frac{3}{4} )}{2 ( \frac{1}{8} )}

Simplify

- \frac{\frac{3}{4}}{2 ( \frac{1}{8} )} : - 3

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 3, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{8}

Minimum (- 3, - 1)

y^{2} - 6 y - 8 x + 17 = 0

y = x^{2} + 3 x - 10

f (x) = x^{2} + 2 x - 15

8 y = x^{2}

x = y^{2} + 3 y + 4

What is the vertices x^2+6x-8y+1=0 ?
The vertices x^2+6x-8y+1=0 is Minimum (-3,-1)