How do you find the vertex of #h(x) + 3/16x^2 - 5/8x - 1/2#?

1 Answer
Feb 28, 2016

#x_("vertex")=5/3#

I have left #y_("vertex")# for you to derive by substitution.

Explanation:

Quick method: Find #x_("vertex")# then substitute it back into the original equation to find #y_("vertex")#

Given:#" "3/16 x^2 -5/8 x-1/2# ............................(1)

'~~~~~~~~~~~~ For reference: ~~~~~~~~~~~~~~~~~
#color(brown)(3/16 xxk =5/8) -> color(green)(k=5/8xx16/3= 10/3)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write equation (1) as:

#3/16(x^2color(green)(-10/3)x)-1/2#

#color(blue)(x_("vertex")= (-1/2)xx(-10/3) =+ 10/6 = +5/3 = 1.6667)#

Tony B