What is the vertices x^2+14x-8y+65=0 ?

The vertices x^2+14x-8y+65=0 is Minimum (-7,2)

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vertices x^2+14x-8y+65=0

vertices

x^{2} + 14 x - 8 y + 65 = 0

Minimum (- 7, 2)

Solution steps

x^{2} + 14 x - 8 y + 65 = 0

Rewrite

x^{2} + 14 x - 8 y + 65 = 0

in the form

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

y = \frac{1 \cdot x ^{2}}{8} + \frac{7 x}{4} + \frac{65}{8}

The parabola parameters are:

a = \frac{1}{8}, b = \frac{7}{4}, c = \frac{65}{8}

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

x_{v} = - \frac{( \frac{7}{4} )}{2 ( \frac{1}{8} )}

Simplify

- \frac{\frac{7}{4}}{2 ( \frac{1}{8} )} : - 7

x_{v} = - 7

Plug in

x_{v} = - 7

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 7, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{8}

Minimum (- 7, 2)

x^{2} - 4 x - 4

f (x) = 2 x^{2} - 6 x + 15

f (x) = 2 x^{2} + 5 x - 3

4 (x + \frac{3}{2})^{2} + 15

f (x) = 2 x^{2} - 8 x

What is the vertices x^2+14x-8y+65=0 ?
The vertices x^2+14x-8y+65=0 is Minimum (-7,2)