How do you solve -3/4x^2+4x-8<0−34x2+4x−8<0 by graphing?

1 Answer
yumean1119
Nov 5, 2017

Graph y=-3/4x^2+4x-8y=−34x2+4x−8 with the vertex, y-intercept and (if any) x-intercepts.

Explanation:

Let y=-3/4x^2+4x-8y=−34x2+4x−8.
[Step1] First, you need to complete the square.
y=-3/4x^2+4x-8y=−34x2+4x−8
y=-3/4(x^2-16/3x)-8y=−34(x2−163x)−8
y=-3/4{(x-8/3)^2-(8/3)^2}-8y=−34{(x−83)2−(83)2}−8
y=-3/4(x-8/3)^2+3/4*64/9-8y=−34(x−83)2+34⋅649−8
y=-3/4(x-8/3)^2-8/3y=−34(x−83)2−83

Its vertex is (8/3,-8/3)(83,−83) and the graph is convex upward.

[Step2] Determine the intercepts.
Put x=0x=0 to the function and yy-intercept is -8−8.
However, there is no xx-interscepts, since yy 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−∞<x<∞, or, any real number.

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