# What is the vertex of y= -2(x - 4)^2 - 5x+3 ?

Jan 6, 2016

The vertex is $\left(\frac{11}{4} , - \frac{111}{8}\right)$

#### Explanation:

One of the forms of the equation of a parabola is $y = a {\left(x - h\right)}^{2} + k$ where (h, k) is the vertex. We can transform the above equation into this format to determine the vertex.

Simplify
$y = - 2 \left({x}^{2} - 8 x + 16\right) - 5 x + 3$

It becomes
$y = - 2 {x}^{2} + 16 x - 32 - 5 x + 3$
$y = - 2 {x}^{2} + 11 x - 29$

Factor out 2 being the coefficient of ${x}^{2}$
$y = - 2 \left({x}^{2} - \frac{11}{2} x + \frac{29}{2}\right)$

Complete the square: Divide by 2 the coefficient of x and then square the result. The resulting value becomes the constant of the perfect square trinomial.

${\left(\frac{- \frac{11}{2}}{2}\right)}^{2} = \frac{121}{16}$

We need to add 121/16 to form a perfect square trinomial. We have to deduct it as well though, to preserve the equality. The equation now becomes

$y = - 2 \left({x}^{2} - \frac{11}{2} x + \frac{121}{16} - \frac{121}{16} + \frac{29}{2}\right)$

Isolate the terms which form the perfect square trinomial

$y = - 2 \left({x}^{2} - \frac{11}{2} x + \frac{121}{16}\right) + \frac{121}{8} - 29$
$y = - 2 \left({x}^{2} - \frac{11}{2} x + \frac{121}{16}\right) - \frac{111}{8}$
$y = - 2 {\left({x}^{2} - \frac{11}{4}\right)}^{2} - \frac{111}{8}$

From this
$h = \frac{11}{4}$
$k = - \frac{111}{8}$
Hence, the vertex is $\left(\frac{11}{4} , - \frac{111}{8}\right)$