How do you graph #f (v ) = \frac { 2v - 9} { 5v + 6}#?

1 Answer
Dec 24, 2017

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Explanation:

Graphing #y=f(v)#:

To graph a rational function (with algebra-only knowledge), we can find key features. These features are

  • v- and y-intercepts (horizontal and vertical axis intercepts)
  • holes
  • asymptotes (vertical, horizontal, slant/oblique)
  • end behavior and behavior around asymptote

v-intercept (horizontal axis intercept)

To find possible v-intercepts, set the function to equal zero to find the zeros.

#f(v) = \frac{2v-9}{5v+6}#
#0 = \frac{2v-9}{5v+6}#

Multiply both sides by 5v+6

#0 = 2v-9#
#9 = 2v#
#v = \frac{9}{2} = 4.5#

So we have an intercept on the v-axis at #(4.5,0)#.

y-intercept (vertical axis intercept)

To find possible y-intercepts, evaluate #f(0)#.

#f(v) = \frac{2v-9}{5v+6}#
#f(0) = \frac{2(0)-9}{5(0)+6}#
#f(0) = \frac{-9}{6}#
#f(0) = \frac{-3}{2} = -1.5#

So we have an intercept on the y-axis at #(0,-1.5)#.

holes

There are no holes. Nothing can be factored or "canceled out."

vertical asymptote

Find what value of #v# results in an undefined answer.

The denominator of #f(v) = \frac{2v-9}{5v+6}# cannot be zero. So the asymptote is at

#5v+6=0 \Leftrightarrow v = -\frac{6}{5} = -1.2#

VA at #v = -1.2#

horizontal asymptote

For this rational function, the degree of the numerator and denominator are the same. So the horizontal asymptote is the ratio of the leading coefficients.

Since #f(v) = \frac{2v-9}{5v+6}#, the leading coefficient of the numerator is 2 and the leading coefficient of the denominator is 5.

HA at #y=\frac{2}{5} = 0.4#

end behavior

Using a calculator, substituting big positive numbers into the formula shows that it is below the horizontal asymptote as #v \to \infty#.

Substituting big negative numbers into the formula shows that it is above the horizontal asymptote as #v \to -\infty#.

behavior around vertical asymptote

Using a calculator, substituting numbers slightly less than (to the left of) the vertical asymptote into the formula shows that #y \to +\infty# as #v \to -1.2^-#.

substituting numbers slightly bigger than (to the right of) the vertical asymptote into the formula shows that #y \to -\infty# as #v \to -1.2^+#.