Rational Expressions and Functions: practice solutions, Grade 11 – download the PDF

1 Finding a domain ★★★

The denominator is zero when \(x^2-25=0\), that is \((x-5)(x+5)=0\), so \(x=5\) or \(x=-5\).

The domain is all real numbers except \(-5\) and \(5\).

2 Simplifying a monomial quotient ★★★

Divide top and bottom by their common factor \(6x^2\): \(\dfrac{12x^3}{18x^2}=\dfrac{2x}{3}\).

The original denominator is zero when \(x=0\), so \(x\neq 0\).

3 Factoring a common factor ★★★

Factor the numerator: \(5x+15=5(x+3)\). So \(\dfrac{5(x+3)}{x+3}=5\) for \(x\neq -3\).

The graph is the horizontal line \(y=5\) with a hole at \((-3,5)\).

4 Multiplying two fractions ★★★

Multiply across: \(\dfrac{12x}{72x^2}\). Divide top and bottom by \(12x\): \(\dfrac{1}{6x}\).

The result is \(\dfrac{1}{6x}\), with \(x\neq 0\).

5 Same denominator ★★★

The denominators match, so add the numerators: \(\dfrac{(2x+1)+(x+7)}{x+4}=\dfrac{3x+8}{x+4}\).

This does not simplify further, and \(x\neq -4\).

6 Reading asymptotes from the form ★★★

The form is \(\dfrac{a}{x-h}+k\) with \(h=-3\) and \(k=-2\).

The vertical asymptote is \(x=-3\) and the horizontal asymptote is \(y=-2\).

7 A cyclist at steady speed ★★★

Direct variation: \(d=kt\). From \(42=k\cdot 3\) we get \(k=14\) miles per hour.

Then \(d=14\cdot 5=70\). She rides 70 miles (about 112.7 km).

8 A trinomial over a difference of squares ★★★

Factor: \(x^2-6x+8=(x-2)(x-4)\) and \(x^2-4=(x-2)(x+2)\).

Restrictions: \(x\neq 2\) and \(x\neq -2\). Cancel \((x-2)\): the result is \(\dfrac{x-4}{x+2}\).

9 Dividing rational expressions ★★★

Multiply by the reciprocal: \(\dfrac{(x-3)(x+3)}{x+2}\cdot\dfrac{(x-2)(x+2)}{x+3}\).

Cancel \((x+3)\) and \((x+2)\): the result is \((x-3)(x-2)\).

Restrictions: \(x\neq -2\) (denominator \(x+2\)), \(x\neq \pm 2\) from \(x^2-4\), and \(x\neq -3\) (the divisor would be \(0\)). So \(x\neq -3,\ -2,\ 2\).

10 Subtracting with different denominators ★★★

The LCD is \((x-2)(x+4)\).

\(\dfrac{5(x+4)-3(x-2)}{(x-2)(x+4)}=\dfrac{5x+20-3x+6}{(x-2)(x+4)}=\dfrac{2x+26}{(x-2)(x+4)}\).

Factoring the top gives \(\dfrac{2(x+13)}{(x-2)(x+4)}\), which cannot be simplified further.

11 A first rational equation ★★★

Multiply both sides by \(x(x-4)\): \(5(x-4)=3x\), so \(5x-20=3x\) and \(2x=20\), \(x=10\).

Check: \(\dfrac{5}{10}=\dfrac12\) and \(\dfrac{3}{6}=\dfrac12\). The solution is \(x=10\).

12 Sketching a rational function ★★★

1. The denominator is zero at \(x=-1\): vertical asymptote \(x=-1\). The degrees are equal and the leading coefficients are 3 and 1: horizontal asymptote \(y=3\).
2. If \(x=0\), then \(f(0)=0\), so the only intercept is the origin \((0,0)\).
3. \(f(2)=\dfrac{6}{3}=2\) and \(f(-3)=\dfrac{-9}{-2}=4.5\). The right branch rises from the origin toward \(y=3\); the left branch lies above \(y=3\) and descends toward it as \(x\to-\infty\).

13 Painters on a fence ★★★

Inverse variation: \(t=\dfrac{k}{n}\), so \(k=nt=8\cdot 9=72\).

With \(n=12\): \(t=\dfrac{72}{12}=6\). Twelve painters need 6 hours.

14 True or false? ★★★

1. False. For \(x=1\): \(\dfrac{6}{8}=0.75\) but \(\dfrac57\approx 0.714\). You cannot cancel terms.
2. True. \(x^2-1=(x-1)(x+1)\), and the factor \(x-1\) cancels whenever \(x\neq 1\).
3. False. For \(x=y=1\): the left side is \(1+1=2\) and the right side is \(\dfrac22=1\).

15 A complex fraction ★★★

Write each part over the LCD \(3x\): the top is \(\dfrac{3+x}{3x}\) and the bottom is \(\dfrac{3-x}{3x}\).

Dividing: \(\dfrac{3+x}{3x}\cdot\dfrac{3x}{3-x}=\dfrac{x+3}{3-x}\).

Restrictions: \(x\neq 0\) (small denominators) and \(x\neq 3\) (the bottom would be \(0\)).

16 Spot the extraneous solution ★★★

Multiply by \(x-4\): \(x-2(x-4)=4\), so \(-x+8=4\) and \(x=4\).

But \(x=4\) makes the denominator \(0\). It is extraneous, so the equation has no solution.

17 A rational equation that becomes a quadratic ★★★

Multiply by \(2x(x+3)\): \(2(x+3)+2x=x(x+3)\), so \(4x+6=x^2+3x\).

Then \(x^2-x-6=0\), that is \((x-3)(x+2)=0\), so \(x=3\) or \(x=-2\).

Check: \(\dfrac13+\dfrac16=\dfrac12\) and \(-\dfrac12+1=\dfrac12\). Both are valid.

18 Two hoses fill a pool ★★★

In one hour, A fills \(\dfrac1t\) of the pool and B fills \(\dfrac{1}{t+6}\). Together: \(\dfrac1t+\dfrac{1}{t+6}=\dfrac14\).

Multiply by \(4t(t+6)\): \(4(t+6)+4t=t(t+6)\), so \(8t+24=t^2+6t\) and \(t^2-2t-24=0\), that is \((t-6)(t+4)=0\).

A time cannot be negative, so \(t=6\). Hose A needs 6 hours and hose B needs 12 hours. Check: \(\dfrac16+\dfrac1{12}=\dfrac14\).

19 Designing a function ★★★

Use the denominator \(x-2\) for the asymptote \(x=2\), and the numerator \(3(x+1)\): the degrees are equal with leading-coefficient ratio \(3\), and the zero is at \(x=-1\).

So \(f(x)=\dfrac{3(x+1)}{x-2}=\dfrac{3x+3}{x-2}\).

The \(y\)-intercept is \(f(0)=\dfrac{3}{-2}=-1.5\).

20 Hole or asymptote? ★★★

1. \(x^2-x-6=(x-3)(x+2)\), so \(g(x)=x+2\) for \(x\neq 3\). The hole is at \(x=3\), height \(3+2=5\): the point \((3,5)\).
2. No: the factor \(x-3\) cancels completely.
3. \(g(0)=\dfrac{-6}{-3}=2\), so the \(y\)-intercept is \((0,2)\). The graph is the line \(y=x+2\) with an open circle at \((3,5)\).

21 Combined variation ★★★

The model is \(y=\dfrac{kx}{z}\). From \(6=\dfrac{4k}{10}\) we get \(k=15\).

Then \(y=\dfrac{15\cdot 7}{5}=21\).