Unit Synthesis — Non-Linear Review
Twenty lessons in one map. Parabolas, circles, hyperbolas, exponentials — match the shape, name the equation, find the features, solve the question.
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Four equations: $y = (x - 3)^2 + 1$, $y = \tfrac{6}{x}$, $x^2 + y^2 = 25$, $y = 2^x$. For each, write (i) the name of the curve, (ii) ONE key feature you'd label on a sketch. Now do the reverse: I show you a U-shape with vertex $(0, -4)$ opening up — what equation am I?
Every non-linear relationship in this unit can be identified by its equation form, sketched by its key features, and analysed by the same toolkit: identify shape $\to$ extract features $\to$ sketch or solve.
Four families. Each has its own signature equation and its own pair of key features. The parabola is the headline act — vertex form, $x$-intercepts, axis of symmetry, transformations, and solving by factoring all live here. Circles, hyperbolas and exponentials round out the family.
Know
- The four key non-linear families and their standard equations
- The key feature checklist for each shape
- The link between solving $y = 0$ and finding $x$-intercepts
Understand
- Why the equation form tells you the shape on sight
- Why transformations of $y = x^2$ produce every parabola
- Why asymptotes mark "not allowed" $x$ or $y$ values
Can Do
- Match any of the four equations to its graph
- Sketch any vertex-form parabola from scratch
- Solve a quadratic by factoring and read the $x$-intercepts off
- Identify any non-linear relationship from a description or sketch
How to identify a curve from its equation in three seconds:
- See $x^2$ only (or $(x - h)^2$): parabola. Look for $a$ to know direction.
- See $x^2 + y^2$ on its own line $= r^2$: circle, centre origin, radius $r$.
- See $x$ in the denominator: hyperbola. Asymptotes on the axes.
- See $x$ in the exponent: exponential. $y$-int $(0, 1)$, asymptote $y = 0$.
- See $y = mx + b$ (no powers, no fractions): linear — not a non-linear relationship.
Parabolas are 60% of this unit. Make sure these five sub-skills are automatic:
- Vertex-form features: $y = a(x - h)^2 + k$ has vertex $(h, k)$, axis $x = h$, $y$-int sub $x = 0$.
- $x$-intercepts: set $y = 0$, isolate $(x - h)^2$, take $\pm$ square root.
- Standard form $\to$ factored: $x^2 + bx + c = (x + p)(x + q)$ with $pq = c$, $p + q = b$.
- Solving: roots of $ax^2 + bx + c = 0$ are the $x$-intercepts.
- Identify (reverse): vertex from sketch + one other point $\to$ solve for $a$ $\to$ equation.
Watch Me Solve It · 3 mixed examples
- 1(a) and (b)(a) Parabola, vertex $(4, 9)$, opens down ($a = -1$). (b) Circle, centre $(0, 0)$, radius $\sqrt{36} = 6$.
- 2(c) and (d)(c) Hyperbola, $k = -2 < 0$, branches in Q2 and Q4, asymptotes the axes. (d) Exponential growth ($a = 3 > 1$), $y$-int $(0, 1)$, asymptote $y = 0$.
- 3Sanity check shapesParabola = U; circle = closed loop; hyperbola = two branches; exponential = J-curve.Match equation form to shape before sketching.
- 1$x$-intercepts$x^2 - 4x - 5 = 0$. $pq = -5$, $p + q = -4$: pick $-5, 1$. $(x - 5)(x + 1) = 0 \Rightarrow x = 5$ or $x = -1$.
- 2$y$-interceptSub $x = 0$: $y = 0 - 0 - 5 = -5$. Point $(0, -5)$.
- 3Axis & vertexAxis: midpoint of $-1$ and $5$ is $2$. So $x = 2$. Vertex $y$: sub $x = 2$ → $y = 4 - 8 - 5 = -9$. Vertex $(2, -9)$.Vertex is a MIN (because $a = 1 > 0$).
- 1Spot the asymptote clueHorizontal asymptote $y = 0$ + $y$-int $(0, 1)$: classic exponential signature.
- 2Increasing means growth$y = a^x$ with $a > 1$ is exponential growth, increasing.
- 3Write a possible equation$y = 2^x$ (or $y = 3^x$, etc.). Any base $> 1$ works.Many equations match the same broad description — only a second point would pin one down.
Common Pitfalls Across the Unit
Parabola
- $y = a(x - h)^2 + k$
- Vertex $(h, k)$, axis $x = h$
- Solve $= 0$ to find $x$-ints
Circle
- $x^2 + y^2 = r^2$ (origin)
- $(x - h)^2 + (y - k)^2 = r^2$
- Centre $(h, k)$, radius $r$
Hyperbola
- $y = \tfrac{k}{x}$
- Asymptotes: both axes
- $k > 0$: Q1, Q3; $k < 0$: Q2, Q4
Exponential
- $y = a^x$, $a > 0$, $a \neq 1$
- $y$-int $(0, 1)$, asymp. $y = 0$
- $a > 1$: growth; $0 < a < 1$: decay
How are you completing this lesson?
Brain Trainer · 4 mixed problems
Four quick identifications spanning all four families.
1 Name the curve $(x - 2)^2 + (y + 1)^2 = 16$ and give its centre and radius.
Both $x^2$ and $y^2$ — circle.Centre $(2, -1)$, radius $4$2 Solve $x^2 - 7x + 10 = 0$ and state the $x$-intercepts of $y = x^2 - 7x + 10$.
$(x - 2)(x - 5) = 0$.$x = 2$ or $5$; ints $(2, 0)$ and $(5, 0)$3 State the vertex and direction of $y = 2(x + 1)^2 - 8$.
$a = 2 > 0$ — opens up.Vertex $(-1, -8)$, MIN4 For $y = \tfrac{4}{x}$, name the shape and which quadrants contain its branches.
$x$ in denominator — hyperbola. $k = 4 > 0$.Branches in Q1 and Q3
Quick Check · 5 mixed questions
Show Your Working · 3 questions
Q6. For each equation, name the curve type and state ONE labelled feature: (a) $y = (x - 1)^2 - 9$, (b) $x^2 + y^2 = 100$, (c) $y = 5^x$.
Q7. Given $y = x^2 - 6x + 8$. (a) Solve $x^2 - 6x + 8 = 0$ by factoring. (b) State the $x$- and $y$-intercepts of the parabola. (c) Find the axis of symmetry and vertex.
Q8. A parabola has vertex $(1, -4)$ and one $x$-intercept at $(3, 0)$. (a) Use the symmetry of the parabola to find the other $x$-intercept. (b) Substitute $(3, 0)$ into $y = a(x - 1)^2 - 4$ to find $a$. (c) Hence write the equation in vertex form AND find the equation in expanded form $y = ax^2 + bx + c$.
Quick Check
1. C — $(x + 2)^2 = (x - (-2))^2$, vertex $(-2, 5)$.
2. B — circle, centre origin, $r = 7$.
3. A — $(x - 3)(x + 4) = 0 \Rightarrow x = 3$ or $-4$.
4. D — $a^0 = 1$, so $(0, 1)$ on all exponentials.
5. B — $y = \tfrac{k}{x}$ has the axes as asymptotes.
Show Your Working Model Answers
Q6 (3 marks): (a) Parabola, vertex $(1, -9)$, opens up [1]. (b) Circle, centre $(0, 0)$, radius $10$ [1]. (c) Exponential growth, $y$-intercept $(0, 1)$, asymptote $y = 0$ [1].
Q7 (3 marks): (a) $pq = 8$, $p + q = -6$: $-2, -4$. $(x - 2)(x - 4) = 0 \Rightarrow x = 2$ or $x = 4$ [1]. (b) $x$-ints $(2, 0)$ and $(4, 0)$; $y$-int $(0, 8)$ [1]. (c) Axis: midpoint of $2$ and $4$ is $3$, so $x = 3$. Vertex: sub $x = 3$ → $y = 9 - 18 + 8 = -1$. Vertex $(3, -1)$ [1].
Q8 (3 marks): (a) Axis $x = 1$. $(3, 0)$ is $2$ right of the axis, so the other intercept is $2$ left: $(-1, 0)$ [1]. (b) $0 = a(3 - 1)^2 - 4 = 4a - 4 \Rightarrow a = 1$ [1]. (c) $y = (x - 1)^2 - 4$. Expand: $y = x^2 - 2x + 1 - 4 = x^2 - 2x - 3$ [1].
What Type Of Relationship Is This?
Five clues. For each, name the curve and write a possible equation. (i) Passes through $(0, -16)$, $x$-intercepts at $\pm 4$. (ii) Closed loop, passes through $(0, 5)$ and $(5, 0)$, centred at origin. (iii) Approaches but never touches $y = 0$, passing through $(0, 1)$ and $(1, 4)$. (iv) Two branches in Q2 and Q4, passing through $(2, -3)$. (v) Parabola with axis $x = -2$, $y$-intercept $(0, -3)$, opening down with $a = -1$.
Reveal solution
(i) Parabola. Roots $\pm 4$, $y$-int $-16$: $y = x^2 - 16$. (ii) Circle, centre $(0, 0)$, radius $5$: $x^2 + y^2 = 25$. (iii) Exponential. $(0, 1)$ always works; $(1, 4)$ gives base $4$: $y = 4^x$. (iv) Hyperbola $y = \tfrac{k}{x}$, $k < 0$. Sub $(2, -3)$: $-3 = \tfrac{k}{2} \Rightarrow k = -6$. So $y = \tfrac{-6}{x}$. (v) $y = -(x + 2)^2 + k$. Sub $(0, -3)$: $-3 = -(2)^2 + k = -4 + k \Rightarrow k = 1$. So $y = -(x + 2)^2 + 1$.
Parabola
$y = a(x - h)^2 + k$
Circle
$x^2 + y^2 = r^2$
Hyperbola
$y = \tfrac{k}{x}$
Exponential
$y = a^x$
Solve
Factor + null factor law
Roots
$=$ $x$-intercepts
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