Question

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

Answer 1

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^+`.