What is the vertex of $y = {(x + 8)}^{2} - 2 x - 6$ $y= (x+8)^2-2x-6$ ?

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Answer 1 · 2017-08-18T16:50:54

See the solution below

$y = x^{2} + 16 x + 64 - 2 x - 6$ $y = x^2 + 16x + 64 -2x -6$
$y = x^{2} + 14 x + 58$ $y = x^2 + 14x + 58$

Since the equation is quadratic, Its graph would be a parabola.
graph{x^2 + 14x + 58 [-42.17, 37.83, -15.52, 24.48]}

As you can see from the graph that the roots are complex for this quadratic equation.

The vertex can be found out by the following formula,
$(x, y) = (- \frac{b}{2 a}, - \frac{D}{4 a})$ $(x,y) = (-b/(2a) , -D/(4a))$

where,
$D =$ $D =$ discriminant

Also
$D = b^{2} - 4 a c$ $D = b^2 - 4ac$

here,
$b = 14$ $b = 14$
$c = 58$ $c = 58$
$a = 1$ $a = 1$

Plugging in the values

$D = 196 - 4 (58) (1)$ $D = 196 - 4(58)(1)$
$D = 196 - 232$ $D = 196 - 232$
$D = - 36$ $D = -36$

Therefore the vertex is given by
$(x, y) = (- \frac{14}{2}, \frac{36}{4})$ $(x,y) = (-14/(2) , 36/4)$
$(x, y) = (- 7, 9)$ $(x,y) = (-7 , 9)$

What is the vertex of y= (x+8)^2-2x-6y=(x+8)2−2x−6 y= (x+8)^2-2x-6?