A factorised quadratic hands you its $x$-intercepts directly. The discriminant tells you how many there are before you factorise anything, and therefore where the parabola sits relative to the axis.
Today's hook, The discriminant you met when solving quadratics has a graphical meaning: it counts how many times the parabola crosses the $x$-axis. Same number, read a different way.
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Recall, your gut answer first
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The quadratic $y = (x - 1)(x - 5)$ is already factorised. Where does it cross the $x$-axis, and how do you know? Then predict what $y = (x-3)^2$ does at $x = 3$.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Zero product, zero height
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Zero product, zero height
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A parabola meets the $x$-axis where $y = 0$. If the quadratic is factorised, the null factor law gives the crossings immediately: $y = (x-1)(x-5)$ is zero when $x = 1$ or $x = 5$. The discriminant counts those crossings without factorising at all.
Factored form $y = a(x - r_1)(x - r_2)$: $x$-intercepts $r_1$ and $r_2$ $\Delta > 0$: two $\Delta = 0$: one $\Delta < 0$: none
Read the roots with the sign flipped
$y = (x - 4)(x + 2)$ has intercepts at $x = 4$ and $x = -2$. The bracket sign is the reverse of the root, exactly as with circles.
$\Delta = 0$ means the parabola touches
One intercept means the vertex sits on the $x$-axis and the curve touches without crossing.
$\Delta < 0$ means it misses entirely
The parabola lies wholly above the axis if $a > 0$, or wholly below if $a < 0$. The sign of $a$ decides which.
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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
In factored form $y = a(x-r_1)(x-r_2)$, the $x$-intercepts are $r_1$ and $r_2$.
The $y$-intercept is found by substituting $x = 0$.
$\Delta = b^2 - 4ac$ counts the $x$-intercepts: two if positive, one if zero, none if negative.
When $\Delta < 0$ the parabola lies entirely on one side of the axis, and the sign of $a$ says which side.
Understand
Concepts
Why the null factor law turns a factorised quadratic straight into its intercepts.
Why $\Delta = 0$ means the curve touches the axis rather than crossing it.
Why $\Delta$ and the sign of $a$ together fix the position of the parabola relative to the axis.
Can do
Skills
Read $x$-intercepts directly from factored form.
Use the discriminant to state the number of $x$-intercepts without solving.
Justify where a parabola sits relative to the $x$-axis from $\Delta$ and the sign of $a$.
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Key terms
ParabolaThe curved graph of a quadratic function. Like this: $y = x^2 - 4$ is a parabola opening upward with its lowest point at $(0, -4)$.
Factored formA quadratic written as a product of brackets, so its roots can be read directly. Like this: $y = (x-1)(x-5)$ has $x$-intercepts 1 and 5.
$x$-intercept of a parabolaA point where the curve meets the $x$-axis, so $y = 0$ there. Like this: $y = (x-4)(x+2)$ meets the axis at $(4,0)$ and $(-2,0)$.
DiscriminantThe expression $b^2 - 4ac$, which counts how many times the parabola crosses the $x$-axis. Like this: $y = x^2 + x + 5$ has $\Delta = -19$, so it never crosses.
Touching the axisWhat a parabola does when it has exactly one $x$-intercept, at its vertex. Like this: $y = (x-3)^2$ touches at $(3,0)$ without passing through.
Concave up and concave downOpening upward when $a > 0$ and downward when $a < 0$, which decides which side of the axis a non-crossing parabola sits. Like this: $y = x^2 + 1$ is concave up and lies entirely above the axis.
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Intercepts from factored form
Work through the core explanation before applying it.
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Intercepts from factored form
core concept
A graph meets the $x$-axis where $y = 0$. For $y = a(x - r_1)(x - r_2)$ that means $a(x-r_1)(x-r_2) = 0$, and by the null factor law either bracket can be the zero.
So the $x$-intercepts are $r_1$ and $r_2$, read straight off with the sign reversed: $y = (x-4)(x+2)$ gives $x = 4$ and $x = -2$.
The $y$-intercept comes from substituting $x = 0$: here $y = (0-4)(0+2) = -8$. Together the three points sketch most of the curve.
A repeated bracket gives one intercept. $y = (x-3)^2$ has both roots at $x = 3$, so the parabola touches the axis at $(3, 0)$ rather than crossing it.
Quick check: what are the $x$-intercepts of $y = (x + 1)(x - 6)$?
In factored form the $x$-intercepts are the values making each bracket zero, so the sign is the reverse of the number in the bracket. The $y$-intercept comes from substituting $x = 0$. A repeated bracket gives a single intercept where the curve touches.
Pause, copy the null factor reasoning, the worked $y = (x-4)(x+2)$ giving intercepts 4 and $-2$ and $y$-intercept $-8$, and the repeated-bracket case, into your book.
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The discriminant counts the crossings
core concept
We just saw that a factorised quadratic hands over its intercepts. That raises a question: what if it does not factorise, or you only need to know how many crossings there are? This card answers it → the discriminant answers the counting question without any factorising.
The $x$-intercepts are the solutions of $ax^2 + bx + c = 0$, so the number of intercepts is the number of real roots, which is exactly what $\Delta = b^2 - 4ac$ reports.
$\Delta > 0$ gives two distinct real roots, so the parabola crosses the axis twice. $\Delta = 0$ gives one, so it touches. $\Delta < 0$ gives none, so it misses the axis entirely.
For $y = x^2 - 6x + 9$: $\Delta = 36 - 36 = 0$, so it touches. Indeed it factorises as $(x-3)^2$, confirming a single intercept at $x = 3$.
Same discriminant, new question. When solving equations $\Delta$ counted solutions. Here it counts crossings. They are the same count, because a crossing is a solution of $y = 0$.
Fill the blank: for $y = x^2 - 6x + 9$ the discriminant is .
The $x$-intercepts are the real roots of $ax^2+bx+c=0$, so $\Delta$ counts them: positive gives two crossings, zero gives one touch, negative gives none. It is the same discriminant as for solving, asked graphically.
Pause, copy the three cases with their graphical meanings, and the worked $y = x^2-6x+9$ giving $\Delta = 0$ and a touch at $x = 3$, into your book.
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Justifying the position of the parabola
core concept
We just saw that $\Delta$ counts crossings. That raises a question: when there are none, is the parabola above the axis or below it? This card answers it → the discriminant alone cannot say, but $\Delta$ together with the sign of $a$ can.
If $\Delta < 0$ the curve never meets the axis, so it must lie entirely on one side. Which side is decided by whether it opens up or down.
$a > 0$ opens upward, so a non-crossing parabola sits **entirely above** the axis, and $y > 0$ for every $x$. $a < 0$ opens downward and sits entirely below, with $y < 0$ for every $x$.
So "show that $y = 2x^2 + 3x + 5$ is always positive" is a two-part answer: $\Delta = 9 - 40 = -31 < 0$ so it never crosses, and $a = 2 > 0$ so it opens upward and therefore lies wholly above the axis.
Both parts are needed. $\Delta < 0$ alone only says the curve misses the axis. Without the sign of $a$ you cannot say which side, and a proof that omits it is incomplete.
Which conclusion does NOT follow for a quadratic with $\Delta < 0$ and $a > 0$?
$\Delta < 0$ means the parabola never meets the axis. The sign of $a$ then says which side: $a > 0$ opens up and lies wholly above, so $y > 0$ always; $a < 0$ lies wholly below. Always-positive and always-negative proofs need both parts.
Pause, copy the two-part justification with the worked $y = 2x^2+3x+5$ shown always positive, and the warning that $\Delta < 0$ alone is not enough, 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.
PROBLEM 1 · INTERCEPTS FROM FACTORED FORM
Find all intercepts of $y = 2(x - 3)(x + 1)$.
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$x$-intercepts: set each bracket to zero, giving $x = 3$ and $x = -1$
The factor 2 cannot be zero, so it does not contribute an intercept.
How many times does $y = 3x^2 - 4x + 2$ cross the $x$-axis?
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$a = 3$, $b = -4$, $c = 2$
Read the coefficients with their signs.
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$\Delta = (-4)^2 - 4(3)(2) = 16 - 24 = -8$
Square $-4$ to get $+16$.
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$\Delta < 0$, so it never crosses the $x$-axis
No real roots means no intercepts.
PROBLEM 3 · PROVING A QUADRATIC IS ALWAYS POSITIVE
Show that $y = 2x^2 + 3x + 5$ is positive for all real values of $x$.
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$\Delta = 3^2 - 4(2)(5) = 9 - 40 = -31$
Calculate the discriminant first.
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$\Delta < 0$, so the parabola never meets the $x$-axis
It lies entirely on one side.
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$a = 2 > 0$, so it opens upward and lies wholly above the axis, giving $y > 0$ for all $x$
Both facts together complete the proof.
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Quick-fire practice
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Quick-fire practice
+10 XP
Find the $x$-intercepts of $y = (x - 7)(x + 3)$.
Find all intercepts of $y = (x - 2)^2$.
How many $x$-intercepts does $y = x^2 + 4x + 4$ have?
Show that $y = -x^2 + 2x - 5$ is negative for all $x$.
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Revisit your prediction
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Revisit your prediction
At the start you predicted what $y = (x-3)^2$ does at $x = 3$. Confirm it using the discriminant, and explain why a repeated factor and a zero discriminant are two ways of saying the same thing.
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Multiple choice
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Multiple choice
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Short answer
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Short answer
ApplyBand 43 marks
Q1. Find all intercepts of $y = 3(x + 2)(x - 4)$ and state whether the parabola opens upward or downward. (3 marks)
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ApplyBand 54 marks
Q2. Show that $y = 4x^2 - 12x + 9$ touches the $x$-axis at exactly one point, and find that point. (4 marks)
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UnderstandBand 43 marks
Q3. Explain why knowing only that $\Delta < 0$ is not enough to conclude that a quadratic is always positive, and state what else is required. (3 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $x = 7$ and $x = -3$. Practice 2: $x$-intercept $x = 2$ only (a touch); $y$-intercept $(0,4)$. Practice 3: $\Delta = 16 - 16 = 0$, so one. Practice 4: $\Delta = 4 - 20 = -16 < 0$ and $a = -1 < 0$, so it opens downward and lies wholly below the axis.
Q1 (3 marks): $x$-intercepts $x = -2$ and $x = 4$ [1]. $y$-intercept: $3(2)(-4) = -24$, so $(0, -24)$ [1]. $a = 3 > 0$, so it opens upward [1].
Q2 (4 marks): $\Delta = (-12)^2 - 4(4)(9) = 144 - 144 = 0$ [1]. A zero discriminant means exactly one $x$-intercept, so the parabola touches the axis [1]. $4x^2 - 12x + 9 = (2x - 3)^2$ [1]. The touch is at $x = \frac{3}{2}$, that is the point $(\frac{3}{2}, 0)$ [1].
Q3 (3 marks): $\Delta < 0$ establishes only that the parabola never meets the $x$-axis, so it lies entirely on one side [1]. It does not say which side [1]. The sign of $a$ is also required: $a > 0$ opens upward and gives $y > 0$ for all $x$, while $a < 0$ opens downward and gives $y < 0$ for all $x$ [1].
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Boss battle · Intercept Hunt
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Read intercepts off factored quadratics and count crossings with the discriminant. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.