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

1 Answer
Jan 6, 2016

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)#