How do you solve #-3/4x^2+4x-8<0# by graphing?

1 Answer
yumean1119
Nov 5, 2017

Graph #y=-3/4x^2+4x-8# with the vertex, y-intercept and (if any) x-intercepts.

Explanation:

Let #y=-3/4x^2+4x-8#.
[Step1] First, you need to complete the square.
#y=-3/4x^2+4x-8#
#y=-3/4(x^2-16/3x)-8#
#y=-3/4{(x-8/3)^2-(8/3)^2}-8#
#y=-3/4(x-8/3)^2+3/4*64/9-8#
#y=-3/4(x-8/3)^2-8/3#

Its vertex is #(8/3,-8/3)# and the graph is convex upward.

[Step2] Determine the intercepts.
Put #x=0# to the function and #y#-intercept is #-8#.
However, there is no #x#-interscepts, since #y# is always
negative.

graph{-3/4(x-8/3)^2-8/3 [-1, 6, -10, 2]}

Therefore, the domain for x that satisfies the given inequation is #-oo< x< oo#, or, any real number.

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