What is the vertex form of #y=-32x^2+80x+2#?

2 Answers
Jan 9, 2017

Vertex form of equation is #y=-32(x^2-5/4)^2+52#

Explanation:

Vertex form of equation is #y=a(x-h)^2+k#

As we have #y=-32x^2+80x+2#

or #y=-32(x^2-80/32x)+2#

or #y=-32(x^2-5/2x)+2#

or #y=-32(x^2-2xx5/4x+(5/4)^2)+2-(-32)xx(5/4)^2#

or #y=-32(x^2-5/4)^2+2+32xx25/16#

or #y=-32(x^2-5/4)^2+2+50#

or #y=-32(x^2-5/4)^2+52#, where vertex is #(-5/4,-48)#
graph{-32x^2+80x+2 [-10, 10, -60, 60]}

Jan 9, 2017

y = - 32(x - 5/4)^2 + 52

Explanation:

#y = - 32x^2 + 80x + 2 #
x-coordinate of vertex:
#x = -b/(2a) = 80/64 = 5/4#
y-coordinate of vertex:
#y(5/4) = -32(25/16) + 80(5/4) + 2 = -50 + 100 + 2 = 52#
Vertex form of y:
#y = - 32(x - 5/4)^2 + 52#