Four features fix a parabola: the two intercepts if they exist, the $y$-intercept, the axis of symmetry and the vertex. Cubics need the same habit applied to three factors instead of two.
Today's hook, Sketching is the fastest check on everything else. If your algebra says the vertex is below the axis but the curve opens up and misses it, one of the two is wrong and the sketch will tell you which.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
+5 XP warm-up
Without calculating anything, say what shape $y = (x-1)(x-2)(x-3)$ has and roughly where it meets the $x$-axis. How many times does it change direction?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Find the features, then draw
Work through the core explanation before applying it.
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Find the features, then draw
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To sketch a parabola: find the $x$-intercepts (if $\Delta \geq 0$), the $y$-intercept, the axis of symmetry $x = -\frac{b}{2a}$, and the vertex. Plot those, then join with a smooth curve opening the way $a$ dictates. A cubic in factored form works the same way, with three crossings instead of two.
Find the axis first, then substitute it back to get the vertex $y$-value. Doing it in that order avoids most arithmetic slips.
Use symmetry to place a fourth point
The $y$-intercept has a mirror image the same distance on the other side of the axis. That fourth point costs nothing and steadies the sketch.
A cubic changes direction at most twice
Three distinct factors means three crossings and two turning points. A repeated factor flattens the curve where it touches.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
A parabola is fixed by its intercepts, axis of symmetry and vertex, plus the sign of $a$.
The $y$-intercept has a mirror image across the axis of symmetry.
To find a parabola from its graph, choose the form that matches the given features and substitute.
Cubics of the form $y = a(x-r_1)(x-r_2)(x-r_3)$ cross at each root; $y = a(x-r)^3 + k$ is a single point of inflection at $(r, k)$.
Understand
Concepts
Why four features are enough to determine a parabola completely.
Why the choice of form (factored, completed-square or general) depends on what the graph shows you.
Why a cubic with three distinct roots must change direction twice.
Can do
Skills
Sketch a parabola showing intercepts, axis of symmetry and vertex.
Find the equation of a parabola from sufficient graphical features.
Sketch cubics in both of the syllabus forms.
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Key terms
Key features of a parabolaThe intercepts, axis of symmetry and vertex, which together fix the curve. Like this: $y = x^2 - 4x + 3$ has intercepts 1 and 3, axis $x = 2$, vertex $(2, -1)$.
Mirror pointThe reflection of a known point across the axis of symmetry, giving an extra plotting point free. Like this: with axis $x = 2$, the $y$-intercept $(0, 3)$ mirrors to $(4, 3)$.
Turning pointA place where a curve changes from rising to falling or the reverse. Like this: a parabola has exactly one; a cubic with three distinct roots has two.
Cubic functionA function whose highest power is 3, so its graph can cross the $x$-axis up to three times. Like this: $y = (x-1)(x-2)(x-3)$ crosses at 1, 2 and 3.
Point of inflectionWhere a curve changes the direction of its bending, which for $y = a(x-r)^3 + k$ sits at $(r,k)$. Like this: $y = (x-2)^3 + 1$ has one at $(2,1)$.
Sufficient featuresEnough information to pin down every unknown in the chosen form. Like this: a vertex and one other point are sufficient for $y = a(x-h)^2+k$.
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Sketching a parabola from its features
Work through the core explanation before applying it.
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Sketching a parabola from its features
core concept
Work in a fixed order. Find the $x$-intercepts by factorising or by the discriminant if they exist; find the $y$-intercept by substituting $x = 0$; find the axis with $x = -\frac{b}{2a}$; then substitute the axis value back to get the vertex.
For $y = x^2 - 4x + 3$: it factorises as $(x-1)(x-3)$, so intercepts at 1 and 3; $y$-intercept 3; axis $x = -\frac{-4}{2} = 2$; vertex $y = 4 - 8 + 3 = -1$, so $(2, -1)$.
Plot those, add the mirror of the $y$-intercept at $(4, 3)$, and join with a smooth curve opening upward because $a > 0$.
Label everything you found. Sketching marks are awarded for the labelled features, not for the smoothness of the curve. An unlabelled correct shape usually scores less than a rough shape with intercepts and vertex marked.
Quick check: what is the axis of symmetry of $y = x^2 - 6x + 5$?
Sketch in order: $x$-intercepts, $y$-intercept, axis $x = -\frac{b}{2a}$, then the vertex by substituting the axis value back. Add the mirror of the $y$-intercept. Open the curve the way the sign of $a$ dictates, and label every feature.
Pause, copy the four-step order, the fully worked $y = x^2-4x+3$ with all features, and the note that marks go to labelled features, into your book.
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Finding a parabola from its graph
core concept
We just saw how to go from equation to sketch. That raises a question: can you run it backwards, building the equation from a picture? This card answers it → yes, by choosing the form that matches the features the graph gives you.
If the graph shows the **vertex** plus one other point, use $y = a(x-h)^2 + k$. Substitute the vertex for $h$ and $k$, then the other point to find $a$.
If it shows the **$x$-intercepts** plus one other point, use $y = a(x-r_1)(x-r_2)$. Substitute the roots, then the other point for $a$.
A parabola with vertex $(2, -1)$ passing through $(0, 3)$: $y = a(x-2)^2 - 1$, and $3 = a(4) - 1$ gives $a = 1$, so $y = (x-2)^2 - 1$.
You always need one extra point. The vertex alone, or the roots alone, leaves $a$ undetermined, and infinitely many parabolas share them. The extra point is what fixes the width.
Fill the blank: a parabola with $x$-intercepts 2 and 5 has the form $y = a(x-2)(x-$ $)$.
Choose the form from what the graph shows: vertex plus a point means $y = a(x-h)^2+k$; roots plus a point means $y = a(x-r_1)(x-r_2)$. Substitute the known features, then use the extra point to find $a$. Without that extra point $a$ is undetermined.
Pause, copy both forms with when to use each, the worked vertex $(2,-1)$ through $(0,3)$ giving $y = (x-2)^2-1$, and the reason an extra point is always needed, into your book.
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Graphing cubics
core concept
We just saw both directions for parabolas. That raises a question: what changes when the highest power is 3? This card answers it → the same habit of finding features first, but with room for three crossings and two turning points.
In factored form $y = a(x-r_1)(x-r_2)(x-r_3)$ the curve crosses at each root. With $a > 0$ it comes up from the bottom left and leaves to the top right, so it must turn twice to get through all three crossings.
For $y = (x-1)(x-2)(x-3)$: crossings at 1, 2 and 3, $y$-intercept $(-1)(-2)(-3) = -6$. Coming from below on the left, it rises through $x=1$, turns down through $x=2$, then turns up through $x=3$.
The other syllabus form is $y = a(x-r)^3 + k$, which does not turn at all. It has a single point of inflection at $(r, k)$ where the bend changes direction, and it flattens there before continuing.
A repeated factor flattens the curve. $y = x(x-2)^2$ crosses at $x = 0$ but only touches at $x = 2$, flattening against the axis, because the repeated factor behaves like the squared bracket in a parabola.
Which is NOT true of $y = (x-1)(x-2)(x-3)$?
Factored cubics cross at each root, and three distinct roots force two turning points. $y = a(x-r)^3+k$ has no turning point, only a point of inflection at $(r,k)$. A repeated factor makes the curve touch and flatten rather than cross.
Pause, copy both cubic forms, the worked $y=(x-1)(x-2)(x-3)$ with its crossings and $y$-intercept, and the repeated-factor flattening, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
$a > 0$, so the curve rises from bottom left to top right, turning twice between the crossings
Three crossings force two turning points.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
Find the axis of symmetry and vertex of $y = x^2 - 8x + 12$.
Find the equation of the parabola with $x$-intercepts $-1$ and 4 passing through $(0, -8)$.
State the $y$-intercept of $y = (x+1)(x-2)(x-4)$.
How many turning points does $y = (x-5)^3 + 2$ have?
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Revisit the cubic you predicted
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Revisit the cubic you predicted
At the start you predicted the shape and crossings of $y = (x-1)(x-2)(x-3)$ and how many times it changes direction. Check each prediction against the worked example, and explain why three distinct crossings force exactly two turning points.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 44 marks
Q1. Sketch $y = x^2 + 2x - 8$, showing the $x$-intercepts, the $y$-intercept, the axis of symmetry and the vertex. (4 marks)
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ApplyBand 53 marks
Q2. A parabola has $x$-intercepts $-2$ and 6 and passes through the point $(0, -24)$. Find its equation in expanded form. (3 marks)
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UnderstandBand 42 marks
Q3. Explain why knowing only the vertex of a parabola is not enough to determine its equation, and state what additional information is needed. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: axis $x = 4$, vertex $(4, -4)$. Practice 2: $y = 2(x+1)(x-4)$, since $-8 = a(1)(-4)$ gives $a = 2$. Practice 3: $(1)(-2)(-4) = 8$. Practice 4: none; it has a point of inflection at $(5, 2)$.
Q1 (4 marks): $x^2 + 2x - 8 = (x+4)(x-2)$, so $x$-intercepts $-4$ and 2 [1]. $y$-intercept $-8$ [1]. Axis $x = -\frac{2}{2} = -1$; vertex $y = 1 - 2 - 8 = -9$, so $(-1, -9)$ [1]. Sketch opening upward with all four features labelled [1].
Q3 (2 marks): Infinitely many parabolas share the same vertex, differing in how wide or narrow they are, which is controlled by $a$ [1]. One further point on the curve is needed, since substituting it into $y = a(x-h)^2+k$ determines $a$ [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Sketch Sprint
earn bronze · silver · gold
Extract key features and sketch parabolas and cubics against the clock. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.