Lesson 9 ~25 min Unit 2 · Non-Linear +85 XP

$x$-Intercepts and $y$-Intercept of Parabolas

Sub $x = 0$ for the $y$-intercept. Sub $y = 0$ for the $x$-intercepts. A parabola can have 0, 1, or 2 $x$-intercepts — learn how to predict which.

Today's hook: $y = x^2 - 4$: where does it cross the $x$-axis? How many crossings are possible for a parabola, and why?

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Think First

warm-up

Three different parabolas: $y = x^2 - 4$, $y = x^2$, and $y = x^2 + 4$. The first has its vertex BELOW the $x$-axis. The second sits ON the $x$-axis. The third sits ABOVE the $x$-axis. All open upward. How many times does each one cross the $x$-axis? Why are the answers different?

Record your answer in your workbook.

The Big Idea

+5 XP

Intercepts are where a curve crosses the axes. The rules are the same as for straight lines:

The red curve $y = x^2 - 4$ has TWO $x$-intercepts at $(\pm 2, 0)$ and a $y$-intercept at $(0, -4)$. The gold curve $y = x^2$ has ONE $x$-intercept — right at the vertex $(0, 0)$. The purple curve $y = x^2 + 4$ never crosses the $x$-axis: NO real $x$-intercepts. All three have $y$-intercepts (you can always sub $x = 0$).

$y$-intercept: sub $x = 0$. $x$-intercepts: sub $y = 0$ and solve.

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

Always exists for a parabola.

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

May give 2, 1, or 0 real solutions.

Predict from vertex

Vertex sign vs opening direction tells you how many crossings.

What You'll Master

objectives

Know

$y$-intercept of $y = f(x)$: substitute $x = 0$
$x$-intercepts of $y = f(x)$: substitute $y = 0$ and solve
A parabola can have 0, 1, or 2 $x$-intercepts

Understand

Why $x$-intercepts of $y = (x - h)^2 + k$ require $-k/a \ge 0$ (or, when $a = 1$, $-k \ge 0$)
Why $x$-intercepts are symmetric about the axis of symmetry
Why the vertex being on the $x$-axis means exactly ONE $x$-intercept

Can Do

Find the $y$-intercept by substitution
Solve $y = 0$ for parabolas in $y = ax^2 + c$ and $y = a(x - h)^2 + k$ form
Predict number of $x$-intercepts before solving

Words You Need

vocabulary

$y$-interceptPoint where the curve crosses the $y$-axis. Found by setting $x = 0$.

$x$-interceptPoint where the curve crosses the $x$-axis. Found by setting $y = 0$.

Root / zeroAnother name for the $x$-intercept — an $x$-value that makes $y = 0$.

Repeated rootWhen the vertex sits ON the $x$-axis: ONE $x$-intercept, counted twice.

No real interceptsWhen the curve stays entirely above (or below) the $x$-axis — no crossings.

Symmetric pairTwo $x$-intercepts equidistant from the axis of symmetry.

Spot the Trap

heads-up

Wrong: $y = x^2 + 4$ crosses the $x$-axis at $(2, 0)$ and $(-2, 0)$.

Right: Setting $y = 0$ gives $x^2 = -4$, which has NO real solutions. Vertex $(0, 4)$ is above the $x$-axis and the curve opens up — it never touches.

Wrong: Forgetting the $\pm$. $x^2 = 9 \Rightarrow x = 3$ only.

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

Finding $x$-Intercepts in Two Forms

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How you solve depends on the form of the equation:

Form 1: $y = ax^2 + c$. Set $y = 0$: $ax^2 + c = 0 \Rightarrow x^2 = -\dfrac{c}{a} \Rightarrow x = \pm\sqrt{-c/a}$ (real iff $-c/a \ge 0$).

Form 2 (vertex form): $y = a(x - h)^2 + k$. Set $y = 0$: $a(x - h)^2 = -k \Rightarrow (x - h)^2 = -\dfrac{k}{a} \Rightarrow x = h \pm \sqrt{-k/a}$ (real iff $-k/a \ge 0$).

The $\pm$ gives you the symmetric pair, automatically equidistant from the axis $x = h$.

Vertex form: $x = h \pm \sqrt{-k/a}$.

Isolate the square

Move the constant before square-rooting.

Always $\pm$

Square-rooting gives a positive AND negative root.

Negative under root

If $-k/a < 0$, no real solutions: no $x$-intercepts.

Predicting How Many Crossings

+5 XP

You can predict the number of $x$-intercepts from the vertex and opening direction — no algebra required.

Vertex on the $x$-axis ($k = 0$): ONE $x$-intercept (a "touch").
Vertex BELOW the axis, opens UP (or vertex ABOVE, opens DOWN): TWO $x$-intercepts.
Vertex ABOVE the axis, opens UP (or vertex BELOW, opens DOWN): NO $x$-intercepts.

Visualise: if the vertex is on the "wrong" side of the $x$-axis for the curve's direction, the curve never reaches the axis.

Sign of $k$ vs sign of $a$ tells you the story.

$k = 0$

Vertex ON the $x$-axis: one repeated root.

$a$, $k$ opposite signs

Two crossings.

$a$, $k$ same sign

No real crossings.

Watch Me Solve It · 3 examples

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

+15 XP per step

PROBLEM

Find the $y$-intercept and the $x$-intercepts of $y = x^2 - 4$.

1
$y$-intercept
Sub $x = 0$: $y = 0 - 4 = -4$. Point $(0, -4)$.
2
$x$-intercepts: set $y = 0$
$0 = x^2 - 4 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
3
Write the points
$x$-intercepts $(2, 0)$ and $(-2, 0)$.
Vertex $(0, -4)$ is below the $x$-axis and curve opens up $\Rightarrow$ two crossings, as predicted.

Answer$y$-intercept $(0, -4)$; $x$-intercepts $(2, 0)$ and $(-2, 0)$.

Watch Me Solve It · Intercepts of $y = (x - 1)^2 - 9$

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PROBLEM

Find the $y$-intercept and the $x$-intercepts of $y = (x - 1)^2 - 9$.

1
$y$-intercept
Sub $x = 0$: $y = (0 - 1)^2 - 9 = 1 - 9 = -8$. Point $(0, -8)$.
2
$x$-intercepts: isolate the square
$0 = (x - 1)^2 - 9 \Rightarrow (x - 1)^2 = 9 \Rightarrow x - 1 = \pm 3$.
3
Solve for $x$
$x = 1 + 3 = 4$ or $x = 1 - 3 = -2$. Points $(4, 0)$ and $(-2, 0)$.
Symmetric about axis $x = 1$: both points are 3 units from the axis.

Answer$y$-intercept $(0, -8)$; $x$-intercepts $(4, 0)$ and $(-2, 0)$.

Watch Me Solve It · No real $x$-intercepts

+15 XP per step

PROBLEM

Show that $y = 2(x - 3)^2 + 5$ has no $x$-intercepts. Find its $y$-intercept.

1
Predict from vertex
Vertex $(3, 5)$ is ABOVE the $x$-axis. Opens UP ($a = 2 > 0$). Same sign $\Rightarrow$ no crossings.
2
Confirm algebraically
$0 = 2(x - 3)^2 + 5 \Rightarrow (x - 3)^2 = -\tfrac{5}{2}$, but a square can't be negative. No real solutions.
3
$y$-intercept
Sub $x = 0$: $y = 2(0 - 3)^2 + 5 = 18 + 5 = 23$. Point $(0, 23)$.
$y$-intercept always exists; $x$-intercepts may not.

AnswerNo real $x$-intercepts; $y$-intercept $(0, 23)$.

Common Pitfalls

heads-up

Reading the wrong intercept off the equation

Saying $x$-intercepts of $y = x^2 - 4$ are at $\pm 4$ (treating $-4$ as the $x$-intercept).

Fix: $-4$ is the $y$-intercept (the constant term). For $x$-intercepts, you must SET $y = 0$ and SOLVE.

Square-rooting one side only

$(x - 1)^2 = 9 \Rightarrow x - 1 = 3$ only, missing the negative root.

Fix: $(x - 1)^2 = 9 \Rightarrow x - 1 = \pm 3$. Always $\pm$ when taking a square root in an equation.

Claiming intercepts exist when they don't

Trying to "solve" $x^2 = -5$ and writing $x = \pm \sqrt{-5}$.

Fix: A square can't be negative for real $x$. State "no real $x$-intercepts" and check it visually using vertex + direction.

Copy Into Your Books

$y$-intercept

Sub $x = 0$
Always one point
Read it off after evaluating

$x$-intercepts

Sub $y = 0$
Isolate the square
Take $\pm$ square root

Vertex form

$0 = a(x - h)^2 + k$
$(x - h)^2 = -k/a$
$x = h \pm \sqrt{-k/a}$

How many?

$k = 0$: one (touch)
$a, k$ opposite signs: two
$a, k$ same signs: none

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Intercepts

4 problems

Four quick problems on intercepts of parabolas.

1 Find the $y$-intercept of $y = (x - 2)^2 + 5$.
Sub $x = 0$: $y = 4 + 5 = 9$.$(0, 9)$
2 Find the $x$-intercepts of $y = x^2 - 16$.
$x^2 = 16 \Rightarrow x = \pm 4$.$(4, 0)$ and $(-4, 0)$
3 How many $x$-intercepts does $y = 3x^2 + 1$ have?
$a = 3 > 0$, $c = 1 > 0$: same sign, no crossings.0 (none)
4 Solve $(x - 4)^2 = 25$ for the $x$-intercepts of $y = (x - 4)^2 - 25$.
$x - 4 = \pm 5 \Rightarrow x = 9$ or $x = -1$.$(9, 0)$ and $(-1, 0)$

Complete in your workbook.

Quick Check · 5 questions

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

+10 XP

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

+10 XP

How many $x$-intercepts does $y = (x - 1)^2 + 4$ have?

+10 XP

The $x$-intercepts of $y = (x - 1)^2 - 9$ are:

+10 XP

How many $x$-intercepts does $y = 3(x - 2)^2$ have?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. For $y = x^2 - 25$, find: (a) the $y$-intercept; (b) the $x$-intercepts (show working).

Answer in your workbook.

ApplyMedium3 MARKS

Q7. Find the $x$-intercepts and $y$-intercept of $y = (x + 2)^2 - 16$. Show all working.

Answer in your workbook.

ReasonHard3 MARKS

Q8. Without solving algebraically, decide how many $x$-intercepts each parabola has. Briefly justify each answer using direction and vertex. (a) $y = -2(x - 1)^2 + 8$; (b) $y = (x + 5)^2$; (c) $y = 4x^2 + 7$.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. A — $y = -4$, point $(0, -4)$.

2. C — $x^2 = 4 \Rightarrow x = \pm 2$.

3. D — vertex above, opens up: none.

4. B — $x = 1 \pm 3$, giving $4$ and $-2$.

5. C — vertex on $x$-axis $\Rightarrow$ one repeated root.

Show Your Working Model Answers

Q6 (3 marks): (a) Sub $x = 0$: $y = -25$. $y$-intercept $(0, -25)$ [1]. (b) Set $y = 0$: $x^2 - 25 = 0 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5$ [1]. Intercepts $(5, 0)$ and $(-5, 0)$ [1].

Q7 (3 marks): $y$-intercept: $y = (0 + 2)^2 - 16 = 4 - 16 = -12$, so $(0, -12)$ [1]. Set $y = 0$: $(x + 2)^2 = 16 \Rightarrow x + 2 = \pm 4$ [1]. $x = 2$ or $x = -6$, so $(2, 0)$ and $(-6, 0)$ [1].

Q8 (3 marks): (a) Opens down ($a = -2$); vertex $(1, 8)$ above $x$-axis: opposite signs $\Rightarrow$ TWO intercepts [1]. (b) Vertex $(-5, 0)$ ON the $x$-axis: ONE repeated intercept [1]. (c) Opens up ($a = 4$); vertex $(0, 7)$ above the $x$-axis: same signs $\Rightarrow$ NONE [1].

Stretch Challenge · +25 XP, +10 coins

Reverse Engineering Intercepts

A parabola in the form $y = (x - h)^2 + k$ has $x$-intercepts at $(1, 0)$ and $(7, 0)$. (a) State the axis of symmetry (it's the midpoint of the two roots). (b) Hence state $h$. (c) Find $k$ by substituting one intercept and solving. (d) Write the equation and find the $y$-intercept.

Reveal solution

(a) Midpoint: $(1 + 7)/2 = 4$. Axis $x = 4$. (b) $h = 4$. (c) Sub $(1, 0)$: $0 = (1 - 4)^2 + k = 9 + k$, so $k = -9$. (d) $y = (x - 4)^2 - 9$. $y$-intercept: sub $x = 0$: $y = 16 - 9 = 7$, so $(0, 7)$.

Quick Review

$y$-intercept

Sub $x = 0$; always exists

$x$-intercepts

Sub $y = 0$; may have 0, 1, or 2

Vertex on $x$-axis

$k = 0$: one repeated root

Two crossings

$a$ and $k$ opposite signs

No crossings

$a$ and $k$ same signs

Vertex form

$x = h \pm \sqrt{-k/a}$

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