Question

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

Answer 1

The vertex is `(11/4, -111/8)`

Explanation:

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

Simplify
`y = -2(x^2 - 8x +16) - 5x + 3`

It becomes
`y = -2x^2+16x-32-5x+3`
`y = -2x^2+11x-29`

Factor out 2 being the coefficient of `x^2`
`y = -2(x^2-11/2x+29/2)`

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.

`((-11/2)/2)^2 = 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(x^2-11/2x+ 121/16 -121/16 +29/2)`

Isolate the terms which form the perfect square trinomial

`y = -2(x^2-11/2x+ 121/16) +121/8 -29`
`y = -2(x^2-11/2x+ 121/16) -111/8`
`y = -2(x^2-11/4)^2 -111/8`

From this
`h = 11/4`
`k = -111/8`
Hence, the vertex is `(11/4, -111/8)`