Lesson 6 ~25 min Unit 2 · Non-Linear +85 XP

Combining: $y = ax^2 + c$

Stretch, flip, slide — all in one equation. $a$ sets the shape; $c$ sets the height of the vertex. Read both, sketch instantly.

Today's hook: $y = 2x^2 - 3$: the $2$ stretches it; the $-3$ drops it. Can you sketch this without a table of values?

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Think First

warm-up

Look at $y = 2x^2 - 3$. The number $2$ in front of $x^2$ tells you something about shape. The number $-3$ at the end tells you something about position. Without computing any $y$-values, what does the graph look like? Where is the vertex? Which way does it open? Is it narrower or wider than $y = x^2$?

Record your answer in your workbook.

The Big Idea

+5 XP

The equation $y = ax^2 + c$ combines L03–L05 in one go. The coefficient $a$ controls shape (direction from its sign, width from its magnitude). The constant $c$ controls vertical position — the vertex sits at $(0, c)$. Read both parts off the equation and you can sketch without a single table value.

The red curve $y = 2x^2 - 3$: $a = 2$ (narrow, opens up), $c = -3$ (vertex at $(0, -3)$). The purple curve $y = -x^2 + 4$: $a = -1$ (standard width, opens down), $c = 4$ (vertex at $(0, 4)$). The gold curve $y = x^2$ is shown for reference at the origin.

$y = ax^2 + c$ — shape from $a$, vertex at $(0, c)$

$a$ does shape

Sign $\to$ direction. Magnitude $\to$ width.

$c$ does height

Vertex at $(0, c)$. Axis still $x = 0$.

Independent jobs

$a$ and $c$ don't interfere — read each separately.

What You'll Master

objectives

Know

$y = ax^2 + c$ has vertex $(0, c)$ and axis $x = 0$
Sign of $a$ gives the direction; $|a|$ gives the width
The sketch needs: direction, width, vertex, and one symmetric pair

Understand

Why $a$ and $c$ act independently on the graph
Why $(0, c)$ is both the vertex and the $y$-intercept
How $a$ and $c$ together determine whether $x$-intercepts exist

Can Do

Sketch $y = 2x^2 - 3$ and $y = -x^2 + 4$ without a full table
Find $x$-intercepts of $y = ax^2 + c$ (when they exist)
Write an equation given direction, width and vertex

Words You Need

vocabulary

General form$y = ax^2 + c$: every parabola in this lesson fits this template.

ShapeThe combination of direction (from sign of $a$) and width (from $|a|$).

Vertex heightThe $y$-coordinate of the vertex — this is just $c$.

Symmetric pairTwo points $(\pm x_0, y_0)$ on the curve — one of each side of the axis $x = 0$.

Quick sketchVertex + direction + one symmetric pair = enough to draw the curve.

Concavity"Concave up" means opens up ($a > 0$); "concave down" means opens down ($a < 0$).

Spot the Trap

heads-up

Wrong: For $y = -x^2 + 4$, vertex is $(0, -4)$. The "$+4$" not the "$-x^2$" sets the vertex height.

Right: Vertex of $y = ax^2 + c$ is $(0, c)$. For $y = -x^2 + 4$ that's $(0, 4)$.

Wrong: $y = 2x^2 - 3$ at $x = 1$ is $(2 \times 1 - 3)^2 = 1$. Order of operations matters: square BEFORE multiplying or subtracting.

Right: $y = 2(1)^2 - 3 = 2 - 3 = -1$. Square, multiply, then add $c$.

Sketching in 4 Steps

+5 XP

You can sketch any $y = ax^2 + c$ in four quick moves:

Direction: sign of $a$ — $+$ opens up, $-$ opens down.
Width: $|a|$ — bigger than $1$ is narrow, smaller is wide.
Vertex: $(0, c)$ — plot it as the lowest (or highest) point.
Anchor pair: pick $x = 1$, compute $y = a(1)^2 + c = a + c$. Plot $(1, a+c)$ and its mirror $(-1, a+c)$.

Smooth U or upside-down U through those three points and you're done.

Direction $\to$ width $\to$ vertex $\to$ anchor pair.

No full table needed

Three points + symmetry beats a seven-point table for sketching.

$x = 1$ is easiest

$x = 1$ gives $y = a + c$ in one step.

Mirror automatically

$(-1, a+c)$ comes free — symmetry across $x = 0$.

$x$-Intercepts of $y = ax^2 + c$

+5 XP

To find $x$-intercepts, set $y = 0$ and solve $ax^2 + c = 0 \Rightarrow x^2 = -\dfrac{c}{a}$.

If $-\dfrac{c}{a} > 0$, two $x$-intercepts: $x = \pm\sqrt{-c/a}$.
If $-\dfrac{c}{a} = 0$, one $x$-intercept (the vertex sits on the $x$-axis).
If $-\dfrac{c}{a} < 0$, no real $x$-intercepts (curve never crosses the $x$-axis).

Example: $y = -x^2 + 4$. $-\dfrac{4}{-1} = 4 > 0$, so $x^2 = 4$, $x = \pm 2$. Intercepts $(\pm 2, 0)$.

$x^2 = -c/a$; real iff vertex sits on the "wrong" side of the $x$-axis from the opening.

Opens up, vertex below

$a > 0, c < 0$: curve definitely cuts the $x$-axis twice.

Opens down, vertex above

$a < 0, c > 0$: also two $x$-intercepts.

Same side $\Rightarrow$ none

$a > 0, c > 0$ (or both negative): no real $x$-intercepts.

Watch Me Solve It · 3 examples

Watch Me Solve It · Sketch $y = 2x^2 - 3$

+15 XP per step

PROBLEM

Sketch $y = 2x^2 - 3$ using direction, width, vertex, and one anchor pair.

1
Direction & width
$a = 2 > 0$ opens up; $|a| = 2 > 1$ narrower than $y = x^2$.
2
Vertex
$(0, c) = (0, -3)$ — this is the minimum.
3
Anchor pair at $x = \pm 1$
$y = 2(1)^2 - 3 = -1$. So $(\pm 1, -1)$.
Draw a smooth narrow U through $(0,-3)$ and the two anchors.

AnswerNarrow upward parabola, vertex $(0, -3)$, passing through $(\pm 1, -1)$.

Watch Me Solve It · Intercepts of $y = -x^2 + 4$

+15 XP per step

PROBLEM

For $y = -x^2 + 4$: (a) state the vertex; (b) find the $x$-intercepts.

1
Read $a$ and $c$
$a = -1$ (opens down), $c = 4$.
2
Vertex
$(0, c) = (0, 4)$ — this is the maximum.
3
Set $y = 0$
$0 = -x^2 + 4 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.

AnswerVertex $(0, 4)$; $x$-intercepts $(2, 0)$ and $(-2, 0)$.

Watch Me Solve It · Equation from features

+15 XP per step

PROBLEM

A parabola opens downward, is narrower than $y = x^2$, and has vertex $(0, 2)$. Write a possible equation.

1
Sign of $a$
Opens down $\Rightarrow$ $a < 0$.
2
Magnitude of $a$
Narrower $\Rightarrow$ $|a| > 1$. Pick e.g. $a = -3$.
3
Value of $c$
Vertex $(0, c) = (0, 2) \Rightarrow c = 2$. Equation: $y = -3x^2 + 2$.

Answer$y = -3x^2 + 2$ (any $a < -1$ works).

Common Pitfalls

heads-up

Reading $c$ off the wrong place

For $y = -2x^2 + 5$, calling the vertex $(0, -2)$ instead of $(0, 5)$.

Fix: $c$ is the CONSTANT term — the number not attached to $x^2$. Here $c = 5$.

Forgetting the negative root

Solving $x^2 = 9$ as $x = 3$ only when finding $x$-intercepts.

Fix: $x^2 = 9 \Rightarrow x = \pm 3$. Parabolas have symmetric $x$-intercepts.

Treating "opens up" as automatic

Assuming every parabola opens upward and missing the negative sign of $a$.

Fix: Always check the sign of the $x^2$ coefficient first. Sign decides direction.

Copy Into Your Books

$y = ax^2 + c$

Vertex $(0, c)$
Axis $x = 0$
Shape from $a$, height from $c$

From $a$

Sign $\to$ direction
$|a|$ $\to$ width
$a = 0$: not a parabola

4-step sketch

Direction
Width
Vertex $(0, c)$
Anchor pair $(\pm 1, a + c)$

$x$-intercepts

$ax^2 + c = 0$
$x = \pm\sqrt{-c/a}$
Real iff $-c/a \ge 0$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Combined Form

4 problems

Four quick problems on $y = ax^2 + c$.

1 State the vertex of $y = 3x^2 - 5$.
$c = -5$.$(0, -5)$
2 Find $y$ at $x = 2$ on $y = -2x^2 + 1$.
$y = -2(4) + 1 = -7$.$y = -7$
3 Describe $y = -\tfrac{1}{2}x^2 + 3$ in one sentence.
$a = -\tfrac{1}{2}$: down, wide. $c = 3$: vertex up at $(0,3)$.Wide downward parabola, vertex $(0, 3)$
4 Find the $x$-intercepts of $y = x^2 - 9$.
$x^2 = 9$.$x = \pm 3$

Complete in your workbook.

Quick Check · 5 questions

The vertex of $y = 2x^2 - 3$ is:

+10 XP

Which best describes $y = -x^2 + 4$?

+10 XP

For $y = 3x^2 - 4$, what is $y$ when $x = 2$?

+10 XP

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

+10 XP

Which equation gives a narrow downward parabola with vertex $(0, 2)$?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. Sketch $y = 2x^2 - 3$ using the 4-step method (direction, width, vertex, anchor pair at $x = \pm 1$). Label the vertex and the anchor points.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. For $y = -x^2 + 9$, find: (a) the vertex; (b) the $y$-intercept; (c) the $x$-intercepts. Show working for (c).

Answer in your workbook.

ReasonHard3 MARKS

Q8. A parabola opens upward, is narrower than $y = x^2$, and has its vertex at $(0, -4)$. (a) Write one possible equation. (b) Find the $x$-intercepts of your equation. (c) Briefly justify why your choice meets each given feature.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — vertex $(0, c) = (0, -3)$.

2. D — opens down ($a = -1$), vertex $(0, 4)$.

3. A — $3(4) - 4 = 8$.

4. B — $x^2 = 4$, $x = \pm 2$.

5. C — $a = -2$ (down, narrow), $c = 2$ (vertex up).

Show Your Working Model Answers

Q6 (3 marks): Direction: opens up ($a = 2 > 0$); width: narrower than $y = x^2$ ($|a| = 2 > 1$) [1]. Vertex $(0, -3)$ labelled [1]. Anchor pair at $x = \pm 1$ gives $y = 2(1)^2 - 3 = -1$, so $(\pm 1, -1)$; smooth narrow U drawn through three labelled points [1].

Q7 (3 marks): (a) Vertex $(0, 9)$ [1]. (b) $y$-intercept $(0, 9)$ [0.5]. (c) Set $y = 0$: $-x^2 + 9 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$, so $(3, 0)$ and $(-3, 0)$ [1.5].

Q8 (3 marks): (a) Example: $y = 3x^2 - 4$ (any $a > 1$, $c = -4$ works) [1]. (b) $3x^2 - 4 = 0 \Rightarrow x^2 = \tfrac{4}{3} \Rightarrow x = \pm \tfrac{2}{\sqrt{3}}$ (approx $\pm 1.15$) [1]. (c) $a = 3 > 0$: opens up; $|a| = 3 > 1$: narrower than $y = x^2$; $c = -4$: vertex $(0, -4)$ [1].

Stretch Challenge · +25 XP, +10 coins

Two Conditions, One Curve

A parabola $y = ax^2 + c$ has vertex $(0, 1)$ and passes through $(2, -7)$. (a) Use the vertex to state $c$. (b) Use the second point to find $a$. (c) Hence write the equation. (d) Does the curve cross the $x$-axis? If so, where?

Reveal solution

(a) Vertex $(0, c) = (0, 1) \Rightarrow c = 1$. (b) $-7 = a(2)^2 + 1 = 4a + 1$, so $4a = -8 \Rightarrow a = -2$. (c) $y = -2x^2 + 1$. (d) Set $y = 0$: $-2x^2 + 1 = 0 \Rightarrow x^2 = \tfrac{1}{2} \Rightarrow x = \pm \tfrac{1}{\sqrt{2}}$ (approx $\pm 0.71$). Two $x$-intercepts.

Quick Review

Equation

$y = ax^2 + c$ — shape + height in one

Vertex

$(0, c)$ — read $c$ off directly

Direction

Sign of $a$: $+$ up, $-$ down

Width

$|a|$: $> 1$ narrow, $< 1$ wide

Sketch

Direction $\to$ width $\to$ vertex $\to$ anchor $(\pm 1, a + c)$

$x$-intercepts

$x = \pm \sqrt{-c/a}$ (real iff $-c/a \ge 0$)

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