OMPT drill — Functions

Rational functions

Fractional functions: asymptotes, excluded values, and solving equations with x in the denominator, where multiplying through without checking creates phantom solutions. Attempt each problem before opening the solution; the notation only becomes automatic by writing it yourself.

Tested inOMPT-A OMPT-B OMPT-D OMPT-F

Lesson

A rational function is a fraction of polynomials, like $f (x) = \frac{2 x + 1}{x - 3}$ , and everything distinctive about it comes from the denominator. Where the denominator hits zero the function simply does not exist, $x = 3$ is banned territory here, and near that banned value the outputs blow up toward plus or minus infinity, drawing a vertical asymptote on the graph. Far away from the action, the function settles toward a horizontal asymptote instead.

Equations with $x$ in the denominator carry a specific danger the OMPT exploits: when you multiply both sides by an expression containing $x$ , you can manufacture a solution that was never really there. The algebra runs clean, the final number looks respectable, and it is wrong: a phantom created by multiplying by zero. The antidote is mechanical and non-negotiable: note the banned values before you start, and compare your answers against them at the end.

Problem 1Numeric answer

The function $f (x) = \frac{3}{x - 5}$ has a vertical asymptote at $x = a$ . What is $a$ ?

Show the worked solution

Answer: 5

A vertical asymptote appears where the denominator hits zero while the numerator does not: $x - 5 = 0$ gives $x = 5$ , and the numerator there is 3. So $a = 5$ . Near that value the outputs blow up — $f (5.01) = 300$ , for instance.

Problem 2Multiple choice

What is the horizontal asymptote of $g (x) = \frac{2 x + 1}{x - 4}$ ?

A $y = 2$
B $y = - 4$
C $y = 0$
D $y = \frac{1}{4}$

Show the worked solution

Answer: A — $y = 2$

Numerator and denominator have the same degree, so for very large $x$ the fraction behaves like the ratio of leading coefficients: $2 x / x = 2$ . The asymptote is $y = 2$ . Trying $x = 1000$ : $2001/996 \approx 2.01$ , already hugging it.

Problem 3Numeric answer

Solve for $w$ : $\frac{12}{w + 2} = 3$ .

Show the worked solution

Answer: 2

I multiply both sides by $w + 2$ : $12 = 3 (w + 2)$ , so $4 = w + 2$ and $w = 2$ . The value does not make the denominator zero ( $2 + 2 = 4$ ), so it is a genuine solution: $12/4 = 3$ .

Problem 4Multiple choice

Simplify $\frac{x ^{2} - 4}{x ^{2} + 2 x}$ as far as possible.

A $\frac{x - 2}{x}$
B $\frac{x + 2}{x}$
C $\frac{- 4}{2 x}$
D $x - 2$

Show the worked solution

Answer: A — $\frac{x - 2}{x}$

I factor top and bottom: $\frac{( x - 2 ) ( x + 2 )}{x ( x + 2 )}$ . The common factor $(x + 2)$ cancels (valid for $x \neq = - 2$ ), leaving $\frac{x - 2}{x}$ . A numeric check at $x = 4$ : original $12/24 = 0.5$ , simplified $2/4 = 0.5$ .

Problem 5Spot the error

A student simplifies $\frac{n ^{2} + 5}{n ^{2}}$ . Step 1: cancel $n^{2}$ above and below. Step 2: result is 5. Step 3: check at $n = 1$ : original gives 6, simplified gives 5 — "close enough". Which step contains the error?

Show the worked solution

Answer: Step 1

Step 1 cancels $n^{2}$ even though it is a term of the numerator, not a factor of it. The honest simplification splits the fraction: $\frac{n ^{2} + 5}{n ^{2}} = 1 + \frac{5}{n ^{2}}$ . Step 3 actually caught the crime — at $n = 1$ the original is 6 and the "simplified" version is 5 — but the student waved it away. When a check fails, the algebra is wrong; there is no "close enough".

Problem 6Numeric answer

A workshop's average cost per unit is $C (q) = \frac{50 + 2 q}{q}$ euros when it produces $q$ units. For which $q$ is the average cost exactly 4 euros?

Show the worked solution

Answer: 25

I set $\frac{50 + 2 q}{q} = 4$ and multiply by $q$ : $50 + 2 q = 4 q$ , so $50 = 2 q$ and $q = 25$ . Checking: $(50 + 50) /25 = 4$ euros per unit, exactly as required.