How do you graph #(6x^2 + 10x - 3) /( 2x + 2)#?

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Answer 1 · 2015-08-12T12:52:16

When plotting graphs, I usually find the axis intercept points, the asymptotes, the stationary points and the points of inflection.

# f(x) = (6x^2+10x-3)/(2x+2) = 3x+2-7/(2x+2) #

To find the axis intercept points solve # f(x) = 0 # and # f(0) #:
# f(x) = 0 #
# 6x^2+10x-3 = 0 #
# x = (-5 pm sqrt(43))/6 #

# f(0) = -3/2 #

Vertical asymptotes (denominator of f(x) = 0):
# 2x+2 = 0 #
# x = -1 #
# lim_(x rarr -1^(+-)) f(x) = ""_+^(-) oo #

No horizontal asymptotes since:
# lim_(x rarr pm oo} f(x) = pm oo #

However:
# g(x) = 3x+2 # is an asymptote since # lim_(x rarr pm oo) -7/(2x+2) = 0^(""^(""_+^-)) #

# lim_(x rarr pm oo} f(x) rArr lim_(x rarr pm oo} g(x)^(""^(""_+^-)) #

Stationary points (first derivative is equal to zero):
# (df)/dx ne 0, x in RR #
So there are no stationary points.

Points of inflection (second derivative is equal to zero):
# (d^2f)/dx^2 ne 0, x in RR #
So there are no points of inflection.

graph{(6x^2+10x-3)/(2x+2) [-5, 5, -25, 25]}