Functions · Book or PDF

Functions 11

Chris Kirkpatrick, Barbara Alldred, Andrew Dmytriw, Shawn Godin, Angelo Lillo, David Pilmer, Susanne Trew, Noel Walker · Grade 11 · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 9 · pp. 550–585

Appendix A: Review of Essential Skills and Knowledge

This appendix refreshes the skills that Grade 11 functions work depends on: arithmetic with integers, fractions and powers, the Pythagorean theorem, lines and linear systems, and the algebra of expressions and quadratic equations. It then moves on to data and curves, parabolas and circles, and trigonometry in right and non-right triangles. This part covers the number skills, linear relations and systems, and algebra with quadratics.

Integer Operations

Adding integers. Compare the signs. If they match, add the sizes and keep the sign: $(-12) + (-5) = -17$ . If they differ, subtract the smaller size from the larger and keep the sign of the number that is larger in size: $18 + (-5) = 13$ . (p. 551)

Subtracting integers. Change the subtraction into adding the opposite. For example, $-15 - (-8)$ becomes $-15 + 8 = -7$ . (p. 551)

Multiplying and dividing integers. The rule is about signs. Matching signs give a positive answer, and different signs give a negative one. So $(-36) \div (-9) = 4$ , but $(-5) \times 9 = -45$ . (p. 551)

More than one operation. Use BEDMAS: brackets, exponents, division and multiplication (left to right), then addition and subtraction (left to right). For $(-6) \times 9 \div 3$ , go left to right: $-54 \div 3 = -18$ . (pp. 551–552)

Rational Numbers

A rational number is a fraction $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$ . (p. 553)

Add or subtract: first rewrite with a common denominator. For $-\frac{2}{5} + \frac{1}{2}$ , use 10: $-\frac{4}{10} + \frac{5}{10} = \frac{1}{10}$ .
Multiply: multiply numerators and denominators, $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$ . Reduce common factors first to keep the numbers small.
Divide: multiply by the reciprocal of the divisor. $\frac{3}{4} \div \left(-\frac{4}{3}\right) = \frac{3}{4} \times \left(-\frac{3}{4}\right) = -\frac{9}{16}$ .

BEDMAS applies here too. (pp. 553–554)

Exponent Laws

In $a^n$ , $a$ is the base and $n$ is the exponent. It counts how many factors of $a$ are multiplied. When the bases are the same: (p. 555)

Multiplying adds exponents: $a^m a^n = a^{m+n}$ . Example: $5^4 \cdot 5^{-3} = 5^1$ .
Dividing subtracts exponents: $\frac{a^m}{a^n} = a^{m-n}$ . Example: $4^6 \div 4^{-2} = 4^8$ .
A power of a power multiplies exponents: $(a^m)^n = a^{mn}$ .

To simplify $3(3^7) \div (3^3)^2$ , write it as $3^8 \div 3^6 = 3^2 = 9$ . (p. 555)

Zero and negative exponents. Dividing a nonzero number by itself gives 1, and the quotient law gives $a^n \div a^n = a^0$ . So $a^0 = 1$ , and $1200^0 = 1$ . A negative exponent means a reciprocal: $a^{-n} = \frac{1}{a^n}$ , so $3^{-2} = \frac{1}{9}$ . (p. 555)

The Pythagorean Theorem

In a right triangle, the hypotenuse is the longest side and lies opposite the right angle. If the legs are $a$ and $b$ and the hypotenuse is $c$ , then $a^2 + b^2 = c^2$ . The area of the square on the hypotenuse equals the combined areas of the squares on the legs. Legs of 6 m and 8 m give $c = \sqrt{36 + 64} = 10$ m. (pp. 557–558)

Graphing Linear Relationships

A linear relation graphs as a straight line, so two known points are enough to draw it. Two common ways to find points are below. (p. 559)

Table of values. Rewrite $2y = 4x - 2$ as $y = 2x - 1$ , then substitute $x$ -values to get $(0, -1)$ , $(1, 1)$ , $(2, 3)$ .
Intercepts. For $2x + 4y = 8$ , set $y = 0$ to get $x = 4$ , and set $x = 0$ to get $y = 2$ . Join $(4, 0)$ and $(0, 2)$ .

Slope and y-intercept. If the equation is $y = mx + b$ , plot $(0, b)$ and then move by the slope, rise over run. For $y = 3x + 4$ , start at $(0, 4)$ and go up 3, right 1, to reach $(1, 7)$ . An equation in another form must first be rearranged into this one. (pp. 559–560)

Linear Systems

A linear system is two or more linear equations considered together. Its solution is the point $(x, y)$ that satisfies all of them, which is where the lines cross. (p. 561)

Graphically. Draw both lines on the same axes. For $y = 2x - 8$ and $x + y = 7$ , the lines meet at $(5, 2)$ . Always verify by substituting into both equations: $2(5) - 8 = 2$ and $5 + 2 = 7$ . Graphs may give only approximate answers, so this check matters. (p. 561)

Substitution. This algebraic method gives exact answers. (p. 562)

Isolate one variable in one equation.
Substitute that expression into the other equation and solve.
Substitute back to find the other variable.

Example: for $3x + 2y + 24 = 0$ and $5y + 2x = -38$ , the first equation gives $y = -12 - \frac{3}{2}x$ . Putting this into the second gives $x = -4$ . Then $y = -12 + 6 = -6$ . The solution is $(-4, -6)$ . (p. 562)

Evaluating Expressions and Formulas

Replace each variable with its value, putting negative values in brackets, then follow BEDMAS. For $2x^2 - y$ with $x = -2$ and $y = 3$ : $2(-2)^2 - 3 = 2(4) - 3 = 5$ . The brackets matter, because the square applies to the negative number. For the cylinder volume $V = \pi r^2 h$ with $r = 2.5$ cm and $h = 7.5$ cm, the result is about 147 cm³. (p. 564)

Simplifying, Expanding and Factoring

Like terms. Only terms with the same variables and exponents combine. $3a - 2b - 5a + b = -2a - b$ . (p. 565)

Distributive property. $a(b + c) = ab + ac$ . Every term inside the bracket gets multiplied: $-4a(2a - 3b) = -8a^2 + 12ab$ . (p. 565)

Two binomials. Multiply each term of the first by each term of the second: $(a + b)(c + d) = ac + ad + bc + bd$ . Then collect like terms. $(2x^2 - 3)(5x^2 + 2) = 10x^4 + 4x^2 - 15x^2 - 6 = 10x^4 - 11x^2 - 6$ . Forgetting one of the four products is the usual mistake. (p. 565)

Factoring undoes expanding. (p. 566)

Common factor first. In $10x^4 - 8x^3 + 6x^5$ , the greatest common factor is $2x^3$ , so the result is $2x^3(5x - 4 + 3x^2)$ .
When $a = 1$ : for $x^2 + bx + c$ , find two numbers with product $c$ and sum $b$ . For $x^2 + 4x - 21$ , the numbers are 7 and $-3$ , so it factors as $(x + 7)(x - 3)$ .
When $a \neq 1$ : look for a common factor, then use decomposition. For $3x^2 + 4x - 4$ , $ac = -12$ , and the numbers 6 and $-2$ add to 4. Split the middle term: $3x^2 + 6x - 2x - 4 = 3x(x + 2) - 2(x + 2) = (3x - 2)(x + 2)$ . Check by expanding.

Solving Quadratic Equations

First write the equation as $ax^2 + bx + c = 0$ . Then either factor and set each factor equal to zero, or use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . (p. 567)

By factoring: $x^2 + 3x = 10$ becomes $x^2 + 3x - 10 = 0$ , so $(x + 5)(x - 2) = 0$ and $x = -5$ or $x = 2$ .
By formula: for $2x^2 - x - 3 = 0$ , $a = 2$ , $b = -1$ , $c = -3$ . Then $x = \frac{1 \pm 5}{4}$ , so $x = \frac{3}{2}$ or $x = -1$ .

Misconception: every quadratic has two roots and can be factored. A quadratic equation can have two roots, one root, or none. Some, such as $2x^2 - x - 1.5 = 0$ , have irrational roots and do not factor over the integers. The quadratic formula still handles them. (p. 567)

Scatter plots, lines of best fit and curves of good fit

A scatter plot graphs pairs of numbers, one point per pair, so you can see a trend. A line of best fit is a straight line that follows the trend. Draw it so it passes through or near as many points as possible, with the leftover points split about evenly above and below it. (p. 569)

The direction and tightness of the trend describe the correlation:

Positive: the points slope up to the right.
Negative: the points slope down to the right.
Strong: the points sit close to a line.
Weak: the points are spread out but still lean along a line.

Example: teen smokers averaged about 16 cigarettes a day in 1981 and about 12 a day by 1996. The points drift downward, so the correlation is negative. (p. 569)

Some data bend. In the golf data, a ball hit at 10 m/s travels 10 m, but at 54 m/s it travels 280 m. Distance grows faster as speed rises, so a straight line fits badly. Sketch a smooth upward-curving curve of good fit instead. (p. 570)

Sketching parabolas from their equations

Standard form. In $y = ax^2 + bx + c$ , the sign of $a$ sets the opening: upward if $a > 0$ , downward if $a < 0$ . A parabola is symmetric, so two points with the same $y$ -value are mirror images. The axis of symmetry lies halfway between them. (p. 572)

Example: $y = -3x^2 - 2x + 7$ has $a = -3$ , so it opens downward.

Factor partially: $y = x(-3x - 2) + 7$ .
Then $y = 7$ when $x = 0$ and when $x = -\frac{2}{3}$ .
The axis of symmetry is halfway between them, at $x = -\frac{1}{3}$ .
Substitute $x = -\frac{1}{3}$ to get $y = 7\frac{1}{3}$ , so the vertex is $\left(-\frac{1}{3}, 7\frac{1}{3}\right)$ . (p. 572)

Vertex form. In $y = a(x - h)^2 + k$ , the vertex is $(h, k)$ . It is a minimum point if $a > 0$ and a maximum point if $a < 0$ . For $y = 2(x + 2)^2 + 3$ , the vertex is $(-2, 3)$ and $a = 2$ , so the minimum value is 3 at $x = -2$ . Substituting $x = 0$ gives $(0, 11)$ . By symmetry, $(-4, 11)$ is also on the curve. (p. 573)

Completing the square

Completing the square turns standard form into vertex form. The steps:

Factor $a$ out of the $x^2$ - and $x$ -terms.
Add and subtract the square of half the $x$ -coefficient inside the bracket.
Group the three terms that make a perfect square.
Multiply the subtracted number by $a$ and move it outside the bracket.
Factor the square and collect the constants. (p. 574)

Example: $y = 2x^2 + 12x - 5$ .

Factor: $y = 2(x^2 + 6x) - 5$ .
Half of 6 is 3, and $3^2 = 9$ , so $y = 2(x^2 + 6x + 9 - 9) - 5$ .
Move the $-9$ out as $-18$ : $y = 2(x + 3)^2 - 18 - 5$ .
Result: $y = 2(x + 3)^2 - 23$ , with vertex $(-3, -23)$ . (p. 574)

Transformations of the base parabola

Every parabola in vertex form is the base curve $y = x^2$ moved or reshaped:

$k$ shifts the curve vertically. Positive $k$ moves it up, negative $k$ moves it down.
$h$ shifts it horizontally, in the direction opposite to its sign. In $(x + 3)^2$ , $h = -3$ , so the curve moves 3 units left.
$a$ stretches, compresses or reflects the curve. If $|a| > 1$ the opening is narrower. If $0 < |a| < 1$ it is wider. If $a < 0$ the curve flips over the x-axis. (pp. 575–576)

Example: $y = -2(x + 3)^2 - 4$ .

Shift the base curve 3 left.
Multiply each $y$ -value by $-2$ , which stretches and reflects it.
Shift 4 down.

The vertex ends up at $(-3, -4)$ . (p. 576)

Circles centred at the origin

Take any point $(x, y)$ on a circle centred at the origin with radius $r$ . Together with the origin and the point $(x, 0)$ , it forms a right triangle with hypotenuse $r$ . The Pythagorean theorem then gives

$x^2 + y^2 = r^2.$

With radius 11, the equation is $x^2 + y^2 = 121$ . If $x^2 + y^2 = 30$ , then $r = \sqrt{30} \approx 5.48$ . (p. 577)

A circle is not a function, because most $x$ -values match two $y$ -values. For $x^2 + y^2 = 25$ and $x = 3$ , we get $y^2 = 16$ , so $y = \pm 4$ . (p. 578)

Right-triangle trigonometry

For an acute angle $A$ in a right triangle:

$\sin A = \frac{\text{opp}}{\text{hyp}}, \quad \cos A = \frac{\text{adj}}{\text{hyp}}, \quad \tan A = \frac{\text{opp}}{\text{adj}}$

"Opposite" and "adjacent" depend on which angle you use. In a 3-4-5 triangle where 3 is opposite $A$ , $\sin A = \frac{3}{5}$ , $\cos A = \frac{4}{5}$ and $\tan A = \frac{3}{4}$ . From the other acute angle, the opposite and adjacent sides trade places. (pp. 579–580)

Finding a side. Pick the ratio that uses the known angle, the known side and the unknown side, then solve. With a $27^\circ$ angle, hypotenuse 35 cm and adjacent side $x$ , use $\cos 27^\circ = \frac{x}{35}$ . Then $x = 35\cos 27^\circ \approx 31$ cm. (p. 581)

Finding an angle. Use the inverse function. A ramp with rise 1 and run 8 has $\tan B = \frac{1}{8}$ , so $B = \tan^{-1}\left(\frac{1}{8}\right) \approx 7^\circ$ . (pp. 580–581)

Sine law and cosine law

The primary ratios only work in right triangles. For other triangles, use these laws. (p. 583)

Sine law: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ . Use it when you know two angles and a side, or two sides and an angle opposite one of them. Example: in triangle XYZ, $X = 84^\circ$ , $Z = 27^\circ$ and $z = 4$ cm. Then $Y = 180^\circ - 111^\circ = 69^\circ$ , and $\frac{y}{\sin 69^\circ} = \frac{4}{\sin 27^\circ}$ gives $y \approx 8.2$ cm. (p. 583)

Cosine law: $c^2 = a^2 + b^2 - 2ab\cos C$ , with matching forms for the other sides. Use it when you know two sides and the contained angle, or all three sides. Example: $y = 17$ cm, $z = 19$ cm and $X = 42^\circ$ . Then $x^2 = 17^2 + 19^2 - 2(17)(19)\cos 42^\circ$ , so $x \approx 13$ cm. The sine law cannot start here, because no side is paired with its opposite angle. (pp. 583–584)

Check yourself

Compute $(-8) + 5$ and $(-6)(-4)$ . State the sign rule you used for each.
Simplify $\frac{x^5 \cdot x^3}{x^4}$ and evaluate $7^0$ and $2^{-3}$ .
Solve $x^2 - 5x - 6 = 0$ by factoring. Then explain why not every quadratic equation has two roots or can be solved by factoring.
Write $y = x^2 - 8x + 3$ in vertex form. State the vertex and whether it is a minimum or a maximum.
A triangle has sides 9 cm and 12 cm with a $50^\circ$ angle between them. Which law finds the third side, and why?

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