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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