Create a function whose graph has the given characteristics. (There is more than one correct answer.)
Vertical asymptote:
Slant asymptote:

Question

Create a function whose graph has the given characteristics. (There is more than one correct answer.) Vertical asymptote: $$ x=5 $$ Slant asymptote: $$ y=-x $$ $$ y= $$

UpStudy AI · Accepted Answer

We wish to construct a rational function $$ f(x)=\frac{P(x)}{Q(x)} $$ that meets the two conditions:

1. It has a vertical asymptote at $$ x=5 $$.
2. It has a slant (oblique) asymptote given by $$ y=-x $$.

Since a vertical asymptote at $$ x=5 $$ is produced when the denominator is zero at $$ x=5 $$, we can choose
$$
Q(x)=x-5.
$$

Next, to have a slant asymptote, the degree of the numerator $$ P(x) $$ must be one more than that of the denominator. Since $$\deg(Q(x))=1$$, we choose $$ P(x) $$ to be quadratic. Write it in the general form
$$
P(x)=-x^2+Cx+D.
$$
The negative sign in front of $$ x^2 $$ is chosen so that when we perform the long division, the quotient polynomial (which represents the asymptote) will start with $$-x$$ (remember that the quotient is the result of dividing $$ P(x) $$ by $$ Q(x) $$ and its leading term comes from dividing the leading terms).

Perform long division (or compare coefficients) by writing
$$
\frac{-x^2+Cx+D}{x-5}= Q(x) + \frac{R(x)}{x-5},
$$
where $$ Q(x) $$ is the quotient that will become the asymptote as $$ x\to \infty $$.

Let the quotient be a linear polynomial,
$$
Q(x)=Ax+B,
$$
and let the remainder be a constant $$ R $$. Then we have
$$
-x^2+Cx+D = (x-5)(Ax+B) + R.
$$

Expanding the product on the right gives:
$$
(x-5)(Ax+B) = Ax^2 + Bx - 5Ax - 5B = Ax^2 + (B-5A)x - 5B.
$$
Thus,
$$
-x^2+Cx+D = Ax^2 + (B-5A)x - 5B + R.
$$

Now, equate the coefficients of like powers of $$ x $$:

1. For $$ x^2 $$: 
$$
A=-1.
$$
2. For $$ x $$:
$$
B-5A = C.
$$
Substitute $$ A=-1 $$ to obtain
$$
B+5 = C \quad \Longrightarrow \quad B = C-5.
$$
3. For the constant term:
$$
-5B + R = D.
$$

The slant asymptote is given by the quotient $$ Q(x)=Ax+B $$. We want it to be
$$
y=-x.
$$
Since we already have $$ A=-1 $$, we require that
$$
B=0.
$$
Substitute $$ B=0 $$ in $$ B=C-5 $$ to get:
$$
C-5=0 \quad \Longrightarrow \quad C=5.
$$

Thus far, we have:
$$
A=-1,\quad C=5,\quad B=0.
$$
The quotient is
$$
Q(x)=-x+0=-x,
$$
as desired.

The remainder $$ R $$ is determined by the relation
$$
-5B+R=D.
$$
Since $$ B=0 $$, this simply becomes
$$
R=D.
$$
Because $$ R $$ is a constant (of degree less than the denominator) its presence does not affect the slant asymptote; indeed, as $$ x\to\infty $$, the term $$ \frac{D}{x-5} $$ tends to zero.

One important point is that the factor $$ (x-5) $$ in the denominator should not cancel with a factor from the numerator. To ensure this, we must choose $$ D \neq 0 $$ so that the numerator, evaluated at $$ x=5 $$, is not zero. (If $$ D $$ were 0, then $$ P(5)= -25+5\cdot 5+0=0 $$ and the cancellation would remove the vertical asymptote at $$ x=5 $$.)

A valid function is, for example, obtained by choosing $$ D=1 $$:
$$
f(x)=\frac{-x^2+5x+1}{x-5}.
$$

This function has:

- A vertical asymptote at $$ x=5 $$ (since the denominator is zero and the numerator is $$ -25+25+1=1 \neq 0 $$ when $$ x=5 $$).
- A slant asymptote $$ y=-x $$ because in the long division,
  $$
  f(x)= -x+\frac{1}{x-5},
  $$
  and the remainder term $$ \frac{1}{x-5} $$ tends to zero as $$ x\to \infty $$.

Thus, one acceptable answer is
$$
y=\frac{-x^2+5x+1}{x-5}.
$$
One possible function is $$ y = \frac{-x^2 + 5x + 1}{x - 5} $$.

Create a function whose graph has the given characteristics. (There is more than one correct answer.)
Vertical asymptote:
Slant asymptote:

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Create a function whose graph has the given characteristics. (There is more than one correct answer.) Vertical asymptote: Slant asymptote:

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