Vertical and Horizontal Asymptotes of Rational Function Practice

The strategy

• Let's first factor the numerator and the denominator so we can find the input values that make them equal to zero.
• The function is undefined for inputs that make the denominator equal to zero.
• Inputs where the function is defined and the numerator is equal to zero are zeros of the function. These are simply the $x$-intercepts of the graph of the function.

• Now we can simplify the function expression by canceling out common factors. Any undefined input that no longer makes the denominator equal to zero is a removable discontinuity. The remaining undefined inputs are vertical asymptotes. 

Factoring the numerator and the denominator

$\dfrac{x^2+4x-32}{x^2-8x+16}=\dfrac{(x+8)(x-4)}{(x-4)(x-4)}$

Identifying discontinuities

We see that the denominator is equal to zero when the input is $x=4$. This means that the function isn't defined for this input. It may either have a vertical asymptote or a removable discontinuity at this $x$-value.

We still need to determine the exact type of discontinuity at this point, but let's determine the zeros of the function first.

Identifying zeros

We see that the numerator is equal to zero when the input is either $x=-8$ or $x=4$.

However, the function doesn't have a zero at $x=4$, because it isn't defined for that value.

Therefore, the function has a zero at $x=-8$.

Simplifying the function

The function can be simplified by canceling out a factor of $(x-4)$:

$f(x)=\dfrac{(x+8)\not{{(x-4)}}}{(x-4)\not {{(x-4)}}}=\dfrac{x+8}{x-4}$, for $x\neq 4$.

Identifying removable discontinuities

The simplified expression, $\dfrac{x+8}{x-4}$, is still undefined for $x=4$.

Therefore, the function has no removable discontinuities.

Identifying vertical asymptotes

Even after simplifying the function expression, $x=4$ makes the denominator equal to zero.

Therefore, the function has a vertical asymptote at $x=4$.

In conclusion, $h$ has:

• a zero at $x=-8$
• a vertical asymptote at $x=4$

The strategy

• Let's first factor the numerator and the denominator so we can find the input values that make them equal to zero.
• The function is undefined for inputs that make the denominator equal to zero.
• Inputs where the function is defined and the numerator is equal to zero are zeros of the function. These are simply the $x$-intercepts of the graph of the function.

• Now we can simplify the function expression by canceling out common factors. Any undefined input that no longer makes the denominator equal to zero is a removable discontinuity. The remaining undefined inputs are vertical asymptotes. 

Factoring the numerator and the denominator

$\dfrac{x^2+6x+8}{x+4}=\dfrac{(x+4)(x+2)}{(x+4)}$

Identifying discontinuities

We see that the denominator is equal to zero when the input is $x=-4$. This means that the function isn't defined for this input. It may either have a vertical asymptote or a removable discontinuity at this $x$-value.

We still need to determine the exact type of discontinuity at this point, but let's determine the zeros of the function first.

Identifying zeros

We see that the numerator is equal to zero when the input is either $x=-4$ or $x=-2$.

However, the function doesn't have a zero at $x=-4$, because it isn't defined for that value.

Therefore, the function has a zero at $x=-2$.

Simplifying the function

The function can be simplified by canceling out a factor of $(x+4)$:

$f(x)=\dfrac{\not{{(x+4)}}(x+2)}{\not{{(x+4)}}}=(x+2)$, for
$x\neq -4$.

Identifying removable discontinuities

The simplified expression, $(x+2)$, is no longer undefined for $x=-4$.

Therefore, the function has a removable discontinuity at $x=-4$.

Identifying vertical asymptotes

After simplifying the function expression, no values make the denominator equal to zero.

Therefore, the function has no vertical asymptotes.

In conclusion, $q$ has:

• a zero at $x=-2$
• a vertical asymptote at $x=-4$

The strategy

• Let's first factor the numerator and the denominator so we can find the input values that make them equal to zero.
• The function is undefined for inputs that make the denominator equal to zero.
• Inputs where the function is defined and the numerator is equal to zero are zeros of the function. These are simply the $x$-intercepts of the graph of the function.

• Now we can simplify the function expression by canceling out common factors. Any undefined input that no longer makes the denominator equal to zero is a removable discontinuity. The remaining undefined inputs are vertical asymptotes.

Therefore, the function has a zero at $x=6$.

Simplifying the function

The function can be simplified by canceling out a factor of $(x+4)$:

$f(x)=\dfrac{\not{{(x+4)}}(x-6)}{(x+6)\not{{(x+4)}}}=\dfrac{x-6}{x+6}$, for $x\neq -4$.

Identifying removable discontinuities

The simplified expression, $\dfrac{x-6}{x+6}$, is no longer undefined for $x=-4$.

Therefore, the function has a removable discontinuity at $x=-4$.

Identifying vertical asymptotes

Even after simplifying the function expression, $x=-6$ makes the denominator equal to zero.

Therefore, the function has a vertical asymptote at $x=-6$.

In conclusion, $f$ has:

a zero at $x=6$ a vertical asymptote at $x=-6$

a removable discontinuity at $x=-4$

The strategy

• Let's first factor the numerator and the denominator so we can find the input values that make them equal to zero.
• The function is undefined for inputs that make the denominator equal to zero.
• Inputs where the function is defined and the numerator is equal to zero are zeros of the function. These are simply the $x$-intercepts of the graph of the function.

• Now we can simplify the function expression by canceling out common factors. Any undefined input that no longer makes the denominator equal to zero is a removable discontinuity. The remaining undefined inputs are vertical asymptotes.

Factoring the numerator and the denominator

$\dfrac{x+8}{x^2-10x+9}=\dfrac{(x+8)}{(x-1)(x-9)}$

Identifying discontinuities

We see that the denominator is equal to zero when the input is either $x=1$ or $x=9$. This means that the function isn't defined for those inputs. It may either have a vertical asymptote or a removable discontinuity at those $x$-values.

We still need to determine the exact type of discontinuity at each point, but let's determine the zeros of the function first.

Identifying zeros

We see that the numerator is equal to zero when the input is $x=-8$.

Therefore, the function has a zero at $x=-8$.

Simplifying the function

The function cannot be simplified further.

Identifying removable discontinuities

Since there are no shared terms between the numerator and denominator, there are no removable discontinuities.

Identifying vertical asymptotes

Since the expression cannot be simplified further, $x=1$ and $x=9$ make the denominator equal to zero.

Therefore, the function has vertical asymptotes at $x=1$ and $x=9$.

In conclusion, $q$ has:

• a zero at $x=-8$
• vertical asymptotes at $x=1$ and $x=9$