Mathematics • Year 9 • Unit 2 • Lesson 20
Unit Synthesis, Identify, Sketch, Solve
Drill the unit's master routine: equation $\to$ shape $\to$ features $\to$ sketch or solve. One worked example pulling four families together. One guided. Eight independent across all the lessons of Unit 2.
1. I do, fully worked example (identify four families)
Read every line. The routine: scan equation $\to$ name family $\to$ state ONE key feature.
Problem. For each equation, name the curve and state ONE key feature: (a) $y = -(x - 4)^2 + 9$, (b) $x^2 + y^2 = 36$, (c) $y = \dfrac{-2}{x}$, (d) $y = 3^x$.
Step 1, (a) $y = -(x - 4)^2 + 9$.
Vertex form $y = a(x - h)^2 + k$ with $a = -1, h = 4, k = 9$. Family: parabola. Feature: vertex $(4, 9)$, opens DOWN ($a < 0$), so vertex is a MAX.
Reason: $(x - 4)^2$ signals shape; $-1$ in front signals direction.
Step 2, (b) $x^2 + y^2 = 36$.
Both $x^2$ AND $y^2$ present, with $= r^2$ form. Family: circle. Feature: centre $(0, 0)$, radius $\sqrt{36} = 6$.
Reason: TWO squared variables together $\Rightarrow$ circle (not a parabola).
Step 3, (c) $y = \dfrac{-2}{x}$.
$x$ in the denominator, $k = -2 < 0$. Family: hyperbola. Feature: branches in Q2 and Q4 (because $k < 0$), asymptotes are both axes ($x = 0$ and $y = 0$).
Reason: positive $k$ gives Q1/Q3; negative $k$ flips to Q2/Q4.
Step 4, (d) $y = 3^x$.
$x$ as the exponent, base $3 > 1$. Family: exponential growth. Feature: $y$-intercept $(0, 1)$ (always, for any base), asymptote $y = 0$.
Reason: $3^0 = 1$ regardless of base, every exponential passes through $(0, 1)$.
Answer: (a) parabola, vertex $(4, 9)$ max; (b) circle, centre $(0, 0)$, radius $6$; (c) hyperbola Q2/Q4; (d) exponential growth, $(0, 1)$.
2. We do, fill in the missing steps (sketch and solve a parabola)
Fill in each blank. 4 marks
Problem. Given $y = x^2 - 4x - 5$: (a) find the $x$-intercepts by factoring, (b) find the $y$-intercept, (c) find the axis of symmetry and vertex.
Step 1, Factor the quadratic. Product $= \_\_\_$, sum $= \_\_\_$. Pick numbers $\_\_\_$ and $\_\_\_$. Factored: $(x \_\_\_)(x \_\_\_) = 0$.
Step 2, $x$-intercepts. By null factor law, $x = \_\_\_$ or $x = \_\_\_$. Coordinate pairs: $(\_\_, 0)$ and $(\_\_, 0)$.
Step 3, $y$-intercept. Sub $x = 0$: $y = 0 - 0 - 5 = \_\_\_$. Point $(0, \_\_)$.
Step 4, Axis and vertex. Axis: midpoint of the two $x$-intercepts $= \dfrac{\_\_ + \_\_}{2} = \_\_$. Vertex $y$: sub $x = $ axis value: $y = \_\_\_ - \_\_\_ - 5 = \_\_\_$. Vertex coordinates: $(\_\_, \_\_)$.
3. You do, independent practice (mixed families)
Show your working. The first four are foundation (identify family, one feature). The middle two are standard (find features). The last two are extension (sketch + solve).
Foundation, name the family, name one feature
3.1 $y = (x - 1)^2 - 9$, name the curve and state ONE feature. 1 mark
3.2 $x^2 + y^2 = 100$, name the curve and state the radius. 1 mark
3.3 $y = 5^x$, name the curve and state the $y$-intercept. 1 mark
3.4 $y = \dfrac{4}{x}$, name the curve and state both asymptotes. 1 mark
Standard, features from the equation
3.5 For $y = 2(x + 1)^2 - 8$: state the vertex, the direction the parabola opens, and whether the vertex is a max or min. 2 marks
3.6 For $(x - 2)^2 + (y + 1)^2 = 16$: state the centre and the radius. 2 marks
Extension, solve and sketch
3.7 For $y = x^2 - 7x + 10$: (a) solve $x^2 - 7x + 10 = 0$ by factoring. (b) State the $x$- and $y$-intercepts. (c) Find the axis of symmetry and vertex. 3 marks
3.8 A scenario is described: a curve passes through $(0, 1)$, is increasing through Quadrant 1, and approaches but never touches $y = 0$ for very negative $x$. (a) Name the family. (b) Suggest ONE possible equation. (c) State why a parabola or hyperbola does NOT match this description. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (faded $y = x^2 - 4x - 5$)
Step 1: product $\mathbf{-5}$, sum $\mathbf{-4}$. Numbers $\mathbf{-5}$ and $\mathbf{1}$. Factored: $(x \mathbf{- 5})(x \mathbf{+ 1}) = 0$.
Step 2: $x = \mathbf{5}$ or $x = \mathbf{-1}$. Pairs: $(\mathbf{5}, 0)$ and $(\mathbf{-1}, 0)$.
Step 3: $y = \mathbf{-5}$. Point $(0, \mathbf{-5})$.
Step 4: axis $= \dfrac{\mathbf{-1} + \mathbf{5}}{2} = \mathbf{2}$. Vertex $y$: sub $x = 2$: $y = \mathbf{4} - \mathbf{8} - 5 = \mathbf{-9}$. Vertex $(\mathbf{2}, \mathbf{-9})$.
3.1, $y = (x - 1)^2 - 9$
Parabola. Feature: vertex $(1, -9)$, opens up ($a = 1 > 0$), MIN.
3.2, $x^2 + y^2 = 100$
Circle, centre $(0, 0)$, radius $\sqrt{100} = 10$.
3.3, $y = 5^x$
Exponential growth ($a = 5 > 1$). $y$-intercept $(0, 1)$ (because $5^0 = 1$).
3.4, $y = \dfrac{4}{x}$
Hyperbola, $k = 4 > 0 \Rightarrow$ branches in Q1 and Q3. Asymptotes: $x = 0$ (vertical) and $y = 0$ (horizontal).
3.5, $y = 2(x + 1)^2 - 8$
Vertex $(-1, -8)$. (Note $(x + 1) = (x - (-1))$, so $h = -1$.) $a = 2 > 0 \Rightarrow$ opens UP. Vertex is a MINIMUM.
3.6, $(x - 2)^2 + (y + 1)^2 = 16$
Circle. Centre $(2, -1)$ (note $(y + 1) = (y - (-1))$, so $k = -1$). Radius $\sqrt{16} = 4$.
3.7, Parabola $y = x^2 - 7x + 10$
(a) Product $10$, sum $-7$: try $-2, -5$. $(x - 2)(x - 5) = 0 \Rightarrow x = 2$ or $x = 5$.
(b) $x$-intercepts: $(2, 0)$ and $(5, 0)$. $y$-intercept (sub $x = 0$): $y = 0 - 0 + 10 = 10$. Point $(0, 10)$.
(c) Axis: midpoint of $2$ and $5$: $\dfrac{2 + 5}{2} = 3.5$. Sub $x = 3.5$: $y = 12.25 - 24.5 + 10 = -2.25$. Vertex $(3.5, -2.25)$.
3.8, Identify from a description
(a) Family: exponential growth.
(b) Possible equation: $y = 2^x$ (or any $y = a^x$ with $a > 1$).
(c) A parabola doesn't match because parabolas have a vertex (a turning point) and either cross the $x$-axis twice, touch it once, or sit entirely above/below, they don't approach the $x$-axis asymptotically. A hyperbola has TWO branches and asymptotes on BOTH axes (not just $y = 0$); a single increasing curve through $(0, 1)$ doesn't fit the hyperbola pattern.