Lesson 10 ~25 min Unit 2 · Non-Linear +85 XP

Sketching and Identifying Parabolas

Five-step sketch: $a$, vertex, $y$-intercept, $x$-intercepts, extra point. Then run it backwards — given a sketch, write the vertex-form equation.

Today's hook: Given a sketch with vertex $(2, -1)$, opening upward, passing through $(0, 3)$: can you write its equation?

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Think First

warm-up

You've now learnt to read every feature of a parabola: $a$ (direction + width), vertex, axis, $x$- and $y$-intercepts. Putting them together gives you a sketch. And running it backwards — given a labelled sketch, can you reconstruct the equation? Take the equation $y = -(x - 2)^2 + 4$. Which features can you read straight off? Now sketch it on plain paper.

Record your answer in your workbook.

The Big Idea

+5 XP

The 5-step sketch turns any equation into a labelled curve in under a minute. Reading a sketch and writing the equation is the same skill in reverse.

The red curve $y = -(x - 2)^2 + 4$: $a = -1$ (down, standard width), vertex $(2, 4)$, $y$-int $(0, 0)$, $x$-ints $(0, 0)$ and $(4, 0)$. Sketched with labels in five quick moves: identify $a$, plot vertex, sub $x = 0$, set $y = 0$, draw smoothly through points.

Sketch: $a \to$ vertex $\to y$-int $\to x$-ints $\to$ extra point.

Five fixed steps

Same recipe every time — no guessing.

Label every point

Markers earn the marks: vertex + intercepts.

Same skill backwards

From sketch, read vertex, sub a point, solve for $a$.

What You'll Master

objectives

Know

The 5-step sketching algorithm (direction/width $\to$ vertex $\to$ $y$-int $\to$ $x$-ints $\to$ extra)
Vertex form $y = a(x - h)^2 + k$ has all five features visible
To write the equation from a sketch, use the vertex for $h$ and $k$, then a second point to find $a$

Understand

Why the vertex + intercepts together give the curve's "skeleton"
Why a second point is needed after the vertex to fix $a$ uniquely
Why symmetric points either side of the axis are free once you know one

Can Do

Sketch any vertex-form parabola with labelled features in five steps
Identify (write the equation) from a labelled sketch with vertex + one extra point
Decide quickly when extra symmetric points are useful

Words You Need

vocabulary

SketchA labelled drawing showing direction, vertex, and intercepts — not a perfect plot.

IdentifyGiven a sketch (or features), write the equation in vertex form.

Extra symmetric pointA point on the curve plus its mirror image across the axis $x = h$.

"Skeleton"Vertex + intercepts: the minimum set of labelled points for a sketch.

Smooth curveParabolas are smooth U/inverted-U shapes — no kinks, no straight bits.

Find $a$Substitute a known point (other than the vertex) into $y = a(x - h)^2 + k$ and solve.

Spot the Trap

heads-up

Wrong: Joining the vertex and the intercepts with straight lines.

Right: A parabola is a SMOOTH curve. Join the points with a continuous U (or inverted U).

Wrong: Given a sketch with vertex $(2, -1)$ passing through $(0, 3)$, writing $y = (x - 2)^2 - 1$ without checking.

Right: Sub $(0, 3)$: $3 = a(0 - 2)^2 - 1 = 4a - 1$, so $a = 1$. In this case $a = 1$ works, but you must always check — another sketch could need $a \neq 1$.

The 5-Step Sketch

+5 XP

Use these five steps in this order for every sketch:

Identify $a$: sign for direction, magnitude for width.
Vertex: read $(h, k)$ from vertex form.
$y$-intercept: sub $x = 0$.
$x$-intercepts: set $y = 0$, isolate the square, take $\pm$.
Extra point (optional): pick an easy $x$ and its symmetric partner.

Plot, label, and connect with a smooth curve.

$a \to (h, k) \to y$-int $\to x$-ints $\to$ extra.

Vertex first

Anchor of the curve — everything fits around it.

Both intercepts

Examiners want intercepts labelled with coordinates.

Smooth and symmetric

Mirror across the axis of symmetry — left half matches the right.

Writing the Equation from a Sketch

+5 XP

To "identify" a parabola from a sketch, you need: the VERTEX (read off the diagram), plus ONE other point on the curve.

Write $y = a(x - h)^2 + k$ with $(h, k)$ filled in from the vertex.
Substitute the other point $(x_1, y_1)$.
Solve the resulting equation for $a$.
Write the full equation with $a$ substituted in.

If only direction (up/down) is given and no second point, you can only state $y = a(x - h)^2 + k$ with a stated sign on $a$ — not a unique equation.

Vertex $\to$ template; second point $\to$ value of $a$.

Pick easy points

$y$-intercepts and $x$-intercepts make $a$ easy to solve for.

$a$ controls steepness

If $a$ comes out negative, the curve opens down — should match the sketch.

Sanity check

Sub another known point into your final equation. Should give the right $y$.

Watch Me Solve It · 3 examples

Watch Me Solve It · Sketch $y = -(x - 2)^2 + 4$

+15 XP per step

PROBLEM

Sketch $y = -(x - 2)^2 + 4$ using all 5 steps. Label vertex and both intercepts.

1
Identify $a$ and vertex
$a = -1$ (opens DOWN, standard width). Vertex $(2, 4)$ — a MAX.
2
$y$-intercept
Sub $x = 0$: $y = -(0 - 2)^2 + 4 = -4 + 4 = 0$. Point $(0, 0)$.
3
$x$-intercepts and sketch
Set $y = 0$: $-(x - 2)^2 + 4 = 0 \Rightarrow (x - 2)^2 = 4 \Rightarrow x - 2 = \pm 2 \Rightarrow x = 0$ or $x = 4$. Smooth inverted U through $(0, 0)$, vertex $(2, 4)$, $(4, 0)$.
$y$-intercept and one $x$-intercept happen to coincide at the origin.

AnswerInverted U, vertex $(2, 4)$ max, $y$-int $(0, 0)$, $x$-ints $(0, 0)$ and $(4, 0)$.

Watch Me Solve It · Equation from a sketch

+15 XP per step

PROBLEM

A parabola has vertex $(2, -1)$, opens upward, and passes through $(0, 3)$. Write its equation in vertex form.

1
Fill in vertex
$y = a(x - 2)^2 + (-1) = a(x - 2)^2 - 1$.
2
Substitute $(0, 3)$
$3 = a(0 - 2)^2 - 1 = 4a - 1$.
3
Solve for $a$ and write equation
$4a = 4 \Rightarrow a = 1$. Equation: $y = (x - 2)^2 - 1$.
$a = 1 > 0$, so it opens up — matches the sketch.

Answer$y = (x - 2)^2 - 1$.

Watch Me Solve It · Identify with stretch

+15 XP per step

PROBLEM

A parabola has vertex $(-1, 5)$, opens downward, and passes through $(1, -3)$. Write its equation in vertex form.

1
Fill in vertex
$y = a(x + 1)^2 + 5$ (flip $h = -1$ into $(x + 1)$).
2
Substitute $(1, -3)$
$-3 = a(1 + 1)^2 + 5 = 4a + 5$.
3
Solve for $a$
$4a = -8 \Rightarrow a = -2$. Equation: $y = -2(x + 1)^2 + 5$.
$a = -2 < 0$ — opens down, matching the sketch. Width: $|a| = 2 > 1$, narrower.

Answer$y = -2(x + 1)^2 + 5$.

Common Pitfalls

heads-up

Forgetting to find $a$

Reading vertex $(2, 3)$ and writing $y = (x - 2)^2 + 3$ without checking a second point.

Fix: ALWAYS substitute one more point to find $a$. Without it, you don't have a unique equation.

Pointy or piecewise sketches

Drawing the parabola as straight-line segments connecting vertex and intercepts.

Fix: A parabola is a SMOOTH curve. Join the points with a flowing U or inverted U, not straight lines.

$a$ sign mismatch

Calculating $a = 2$ algebraically but the sketch clearly opens downward.

Fix: Check that the sign of your computed $a$ matches the sketch. If they disagree, re-check the substitution.

Copy Into Your Books

5-step sketch

$a$ (direction, width)
Vertex $(h, k)$
$y$-int (sub $x = 0$)
$x$-ints (sub $y = 0$)
Extra point + label

From sketch to equation

$y = a(x - h)^2 + k$
Use vertex for $h, k$
Sub second point
Solve for $a$

Check your $a$

$a > 0$: opens up
$a < 0$: opens down
Sign must match sketch

Curve drawing tips

Smooth, not pointy
Symmetric L/R about axis
Label ALL key points

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Sketch & Identify

4 problems

Four quick problems mixing sketch features and equation building.

1 State vertex and direction of $y = -(x - 3)^2 + 2$.
$a = -1$, $h = 3$, $k = 2$.Vertex $(3, 2)$, opens down
2 Find $y$-intercept of $y = 2(x - 1)^2 + 3$.
$y = 2(0 - 1)^2 + 3 = 2 + 3$.$(0, 5)$
3 A parabola has vertex $(0, 0)$ and passes through $(2, 12)$. Find $a$.
$12 = a(2)^2 = 4a$.$a = 3$ (equation $y = 3x^2$)
4 Vertex $(1, -4)$, passes through $(3, 0)$ — find $a$ and write the equation.
$0 = a(3 - 1)^2 - 4 = 4a - 4$, so $a = 1$.$y = (x - 1)^2 - 4$

Complete in your workbook.

Quick Check · 5 questions

The $y$-intercept of $y = -(x - 2)^2 + 4$ is:

+10 XP

The $x$-intercepts of $y = -(x - 2)^2 + 4$ are:

+10 XP

A parabola with vertex $(2, -1)$ passes through $(0, 3)$. Its equation is:

+10 XP

A parabola has vertex at the origin and passes through $(2, 12)$. The value of $a$ is:

+10 XP

To write the unique equation of a parabola in vertex form, you need:

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. Sketch $y = (x - 1)^2 - 4$. Label the vertex, $y$-intercept and both $x$-intercepts with coordinates.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. A parabola has vertex $(3, 2)$ and passes through $(5, 10)$. Write its equation in vertex form. Show every step.

Answer in your workbook.

ReasonHard3 MARKS

Q8. A downward-opening parabola has $x$-intercepts at $(-2, 0)$ and $(6, 0)$, and a maximum value of $y = 8$. (a) State the axis of symmetry using the midpoint of the $x$-intercepts. (b) Hence write the vertex. (c) Substitute one $x$-intercept to find $a$, and write the equation in vertex form.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. D — $y = -(0 - 2)^2 + 4 = 0$.

2. B — $(x - 2)^2 = 4 \Rightarrow x = 0$ or $4$.

3. A — $a = 1$, $y = (x - 2)^2 - 1$.

4. C — $12 = 4a \Rightarrow a = 3$.

5. D — vertex fixes $h, k$; second point fixes $a$.

Show Your Working Model Answers

Q6 (3 marks): Vertex $(1, -4)$, opens up [1]. $y$-int: $y = (0 - 1)^2 - 4 = -3$, point $(0, -3)$ [1]. $x$-ints: $(x - 1)^2 = 4 \Rightarrow x - 1 = \pm 2 \Rightarrow x = -1$ or $3$. Smooth U through $(-1, 0)$, $(0, -3)$, $(1, -4)$, $(3, 0)$ [1].

Q7 (3 marks): Template: $y = a(x - 3)^2 + 2$ [1]. Sub $(5, 10)$: $10 = a(5 - 3)^2 + 2 = 4a + 2 \Rightarrow 4a = 8 \Rightarrow a = 2$ [1]. Equation: $y = 2(x - 3)^2 + 2$ [1].

Q8 (3 marks): (a) Midpoint of $-2$ and $6$ is $\tfrac{-2 + 6}{2} = 2$. Axis $x = 2$ [1]. (b) Maximum is at vertex; max $y = 8$ gives vertex $(2, 8)$ [1]. (c) Sub $(-2, 0)$ into $y = a(x - 2)^2 + 8$: $0 = a(-4)^2 + 8 = 16a + 8 \Rightarrow a = -\tfrac{1}{2}$. Equation: $y = -\tfrac{1}{2}(x - 2)^2 + 8$ [1].

Stretch Challenge · +25 XP, +10 coins

Three Points Define a Parabola

A parabola passes through $(1, 0)$, $(5, 0)$ (so these are the $x$-intercepts) and $(3, -8)$. (a) Use the symmetry of the $x$-intercepts to state the axis. (b) Hence state $h$. (c) Substitute $(3, -8)$ to find $k$, knowing $(3, -8)$ is the vertex. Wait — check: is $(3, -8)$ the vertex? Use the axis to justify, then find $a$ from one $x$-intercept. (d) Write the equation in vertex form.

Reveal solution

(a) Midpoint of $1$ and $5$: $3$. Axis $x = 3$. (b) $h = 3$. (c) The point $(3, -8)$ sits on the axis of symmetry, so it must be the vertex. $k = -8$. (d) Template $y = a(x - 3)^2 - 8$. Sub $(1, 0)$: $0 = a(1 - 3)^2 - 8 = 4a - 8 \Rightarrow a = 2$. Equation: $y = 2(x - 3)^2 - 8$.

Quick Review

Step 1

Read $a$: direction + width

Step 2

Plot vertex $(h, k)$

Step 3

$y$-intercept: sub $x = 0$

Step 4

$x$-intercepts: set $y = 0$

Step 5

Extra point + smooth curve

Reverse

Vertex + 1 point $\to$ solve for $a$

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