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.