Quadratics & Cubics — The Crash Course

Section 01

The three forms of a quadratic

Every quadratic is the same animal in three different outfits. Knowing which outfit is which — and how to swap between them — is the foundation of the whole topic.

A quadratic function is any polynomial where the highest power of $x$ is $2$. The graph is always a parabola, which is concave up if $a > 0$ and concave down if $a < 0$.

General form

$y = ax^2 + bx + c$

Best for finding the $y$-intercept (it's just $c$) and for using the axis-of-symmetry shortcut $x = -\frac{b}{2a}$.

Factored form

$y = a(x-\alpha)(x-\beta)$

Best for reading $x$-intercepts directly: they are $\alpha$ and $\beta$. The axis of symmetry sits halfway between them.

Vertex form

$y = a(x-h)^2 + k$

Best for reading the vertex directly: it sits at $(h, k)$. The sign trick: $(x+2)^2$ means $h = -2$, not $+2$.

Sign trap In vertex form $y = a(x-h)^2 + k$, the $h$ is what you subtract. So $y = (x+3)^2 - 5$ has vertex $(-3, -5)$, not $(3, -5)$.

Rewrite $y = 2x^2 + 10x + 12$ in factored form and vertex form.

Factored: Take out the $2$, then factorise.

$y = 2(x^2 + 5x + 6) = 2(x+2)(x+3)$

Vertex: Complete the square.

$y = 2(x^2 + 5x) + 12 = 2\left[(x + \tfrac{5}{2})^2 - \tfrac{25}{4}\right] + 12 = 2(x + \tfrac{5}{2})^2 - \tfrac{1}{2}$

Vertex is $\left(-\tfrac{5}{2}, -\tfrac{1}{2}\right)$.

Section 01 · Three Forms

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QUESTION 01

Which form is best for reading the $y$-intercept straight off the page?

Why: In general form, the constant $c$ is the $y$-intercept — substituting $x = 0$ gives $y = c$ immediately.

QUESTION 02

What is the vertex of $y = 3(x - 4)^2 + 5$?

Why: Vertex form is $y = a(x-h)^2 + k$ with vertex $(h, k)$. Here $(x-4)$ gives $h = 4$ — the sign flips when you read it.

QUESTION 03

What are the $x$-intercepts of $y = -2(x + 3)(x - 1)$?

Why: Each factor equals zero at a root. $(x+3) = 0$ gives $x = -3$, and $(x-1) = 0$ gives $x = 1$.

QUESTION 04

What is the vertex of $y = (x + 2)^2 - 7$?

Why: $(x+2)^2 = (x-(-2))^2$, so $h = -2$. The constant $-7$ is $k$. Vertex is $(-2, -7)$.

QUESTION 05

Expanding $y = -(x - 1)(x + 5)$ gives:

Why: $(x-1)(x+5) = x^2 + 4x - 5$. Negating: $-x^2 - 4x + 5$.

Section 02

The discriminant

The single most overpowered formula in this topic. It tells you how many times your parabola hits the x-axis — without you ever needing to solve the equation.

The discriminant $\Delta = b^2 - 4ac$

It's literally the bit under the square-root sign of the quadratic formula. Three things can happen:

$\Delta > 0$

Two distinct roots

Cuts the x-axis twice. Perfect square → rational. Otherwise → irrational.

$\Delta = 0$

Equal / double root

Touches the x-axis once. The vertex sits exactly on the axis.

$\Delta < 0$

No real roots

Doesn't cross at all. Positive definite if $a>0$, negative definite if $a<0$.

Watch the discriminant in motion

The parabola $y = x^2 + c$ slides up and down as $c$ changes. Watch $\Delta = -4c$ shift through positive, zero, and negative — and watch the roots appear, merge, then vanish.

Constant c = 0.00

Discriminant Δ 0.00

Real roots One

Definite parabolas A parabola is positive definite if it sits entirely above the x-axis (always positive). This requires $\Delta < 0$ AND $a > 0$. Negative definite is the mirror: $\Delta < 0$ AND $a < 0$. Both conditions are needed — don't forget the second one.

Finding values of $k$

A huge chunk of discriminant questions ask you to find values of a parameter $k$ that make a certain root-condition happen. The recipe is always the same:

Identify $a$, $b$, $c$ in terms of $k$.
Write $\Delta = b^2 - 4ac$ and simplify into a quadratic (or inequality) in $k$.
Apply the condition: $\Delta > 0$, $= 0$, or $< 0$.
Solve for $k$ — often this itself is a quadratic inequality.

Find $k$ such that $x^2 - (k+4)x + (k+7) = 0$ has equal roots.

Equal roots means $\Delta = 0$. Here $a = 1$, $b = -(k+4)$, $c = k+7$.

$\Delta = (k+4)^2 - 4(k+7) = k^2 + 8k + 16 - 4k - 28 = k^2 + 4k - 12 = 0$

$(k+6)(k-2) = 0$, so $k = -6$ or $k = 2$.

Watch the coefficient of $x^2$ If the question says "the equation" and the coefficient of $x^2$ contains $k$ (e.g. $(k+1)x^2 - kx + 1 = 0$), then $k = -1$ would make it linear, not quadratic. State this case separately or exclude it.

Section 02 · Discriminant

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QUESTION 01

For $y = x^2 - 4x + 3$, calculate $\Delta$.

Why: $\Delta = b^2 - 4ac = (-4)^2 - 4(1)(3) = 16 - 12 = 4$.

QUESTION 02

How many real roots does $y = 2x^2 - 4x + 2$ have?

Why: $\Delta = (-4)^2 - 4(2)(2) = 16 - 16 = 0$. The parabola touches the $x$-axis at exactly one point.

QUESTION 03

For what values of $k$ does $x^2 + kx + 9 = 0$ have equal roots?

Why: Equal roots ⇒ $\Delta = 0$. So $k^2 - 36 = 0$, giving $k^2 = 36$ and $k = \pm 6$.

QUESTION 04

A parabola is positive definite when:

Why: Positive definite means the parabola sits entirely above the $x$-axis. That needs no real roots ($\Delta < 0$) AND concave up ($a > 0$).

QUESTION 05

The roots of $y = x^2 + 5x + 6$ are:

Why: $\Delta = 25 - 24 = 1$. Positive and a perfect square ⇒ two distinct rational roots. (It factors as $(x+2)(x+3)$.)

Section 03

Parabola anatomy

Five features, every time: concavity, y-intercept, x-intercepts, axis of symmetry, vertex. Lock these down and you can sketch any parabola in under a minute.

The vertex and axis on the move

Here the parabola is $y = (x-h)^2 + k$ with $h$ and $k$ drifting. Notice how the vertex and axis of symmetry move together — the axis is always $x = h$.

Vertex (0.0, 0.0)

Axis of symmetry x = 0.00

Concavity Up (a = 1)

The axis of symmetry

If a parabola has two x-intercepts, the axis sits exactly halfway between them. If not, use the formula — which comes from completing the square on the general form:

Axis of symmetry $x = -\dfrac{b}{2a}$

Finding the vertex

The vertex lies on the axis of symmetry, so its x-coordinate is $-\frac{b}{2a}$. To get the y-coordinate, substitute that x-value back into the original equation.

Domain & Range

Every parabola has domain $(-\infty, \infty)$. The range depends on concavity and the y-value of the vertex (call it $k$):

If $a > 0$ (concave up, minimum at vertex): range is $[k, \infty)$
If $a < 0$ (concave down, maximum at vertex): range is $(-\infty, k]$

Find the vertex of $f(x) = -x^2 + 2x + 8$.

$a = -1, b = 2$, so axis of symmetry: $x = -\dfrac{2}{2(-1)} = 1$.

Sub $x = 1$: $f(1) = -1 + 2 + 8 = 9$. Vertex is $(1, 9)$.

Since $a < 0$, this is a maximum. Range is $(-\infty, 9]$.

Section 03 · Parabola Anatomy

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QUESTION 01

What is the axis of symmetry of $y = 2x^2 - 8x + 5$?

Why: $x = -\dfrac{b}{2a} = -\dfrac{-8}{2(2)} = \dfrac{8}{4} = 2$.

QUESTION 02

Find the vertex of $y = x^2 - 6x + 11$.

Why: Axis: $x = -\dfrac{-6}{2} = 3$. Sub $x = 3$: $y = 9 - 18 + 11 = 2$. Vertex $(3, 2)$.

QUESTION 03

What is the range of $y = -x^2 + 4x - 1$?

Why: $a < 0$ means concave down (maximum). Vertex: $x = 2$, $y = -4 + 8 - 1 = 3$. So $y$ never exceeds $3$, giving range $(-\infty, 3]$.

QUESTION 04

What is the domain of every quadratic function?

Why: You can substitute any real $x$ into $ax^2 + bx + c$. The range varies, but the domain is always $\mathbb{R}$.

QUESTION 05

If a parabola has $x$-intercepts at $-3$ and $7$, where is its axis of symmetry?

Why: The axis sits halfway between the roots: $\dfrac{-3 + 7}{2} = 2$.

Section 04

Sketching parabolas

A "sketch" is not a "graph". You need the key features labelled — not graph-paper accuracy. The form you're given determines the easiest route.

Sketching checklist Every sketch needs: concavity, y-intercept, x-intercepts (if they exist), and the vertex. Label each one. That's it.

From general form $y = ax^2 + bx + c$

Concavity from sign of $a$.
y-intercept is $c$ (or sub $x = 0$).
x-intercepts: factor or use the quadratic formula.
Axis of symmetry: $x = -\frac{b}{2a}$.
Vertex y-value: sub the axis into the equation.

From factored form $y = a(x - \alpha)(x - \beta)$

Concavity from $a$.
x-intercepts straight off: $\alpha$ and $\beta$.
Axis of symmetry: $x = \frac{\alpha + \beta}{2}$.
y-intercept: sub $x = 0$.
Vertex y-value: sub the axis into the equation.

From vertex form $y = a(x - h)^2 + k$

Concavity from $a$.
Vertex is $(h, k)$ — straight off the page.
y-intercept: sub $x = 0$.
x-intercepts: solve $a(x-h)^2 + k = 0$ if they exist.

Sketch $y = (2-x)(x-3)$.

Concavity: Expanding gives $y = -x^2 + 5x - 6$, so $a = -1$ → concave down.

x-intercepts: $x = 2$ and $x = 3$.

y-intercept: $y = (2)(-3) = -6$.

Axis of symmetry: $x = \frac{2+3}{2} = \frac{5}{2}$.

Vertex: $y = -\left(\tfrac{5}{2}\right)^2 + 5\left(\tfrac{5}{2}\right) - 6 = \tfrac{1}{4}$. Vertex is $\left(\tfrac{5}{2}, \tfrac{1}{4}\right)$.

Section 04 · Sketching Parabolas

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QUESTION 01

For $y = -(x-2)^2 + 4$, what is the concavity?

Why: $a = -1 < 0$ — that's a concave-down parabola. The "$-$" out the front does the work.

QUESTION 02

The $y$-intercept of $y = 3x^2 - 5x + 7$ is:

Why: Substitute $x = 0$: $y = 7$. In general form, the $y$-intercept is just $c$.

QUESTION 03

For $y = (x+1)(x-5)$, the axis of symmetry is:

Why: Roots are $-1$ and $5$. Halfway between: $\dfrac{-1+5}{2} = 2$.

QUESTION 04

The vertex of $y = 2(x + 3)^2 - 8$ is:

Why: $(x+3)^2 = (x-(-3))^2$, so $h = -3$, and the constant $-8$ is $k$. Vertex $(-3, -8)$.

QUESTION 05

Which feature is NOT essential for a complete parabola sketch?

Why: Focus/directrix isn't required at Year 11. Vertex, $y$-intercept, $x$-intercepts and concavity are the four things to label.

Section 05

Finding the equation

Given a few features of a parabola, work backwards to its equation. The trick: pick the form that matches what you've been given.

You have: x-intercepts + a

Use factored form

$y = a(x - \alpha)(x - \beta)$. Plug in $\alpha$, $\beta$ and your known $a$, then expand if asked for general form.

You have: vertex + one point

Use vertex form

$y = a(x - h)^2 + k$. Plug in the vertex for $(h, k)$, then sub the extra point to solve for $a$.

You have: three points, no other info

Use simultaneous equations

Start with $y = ax^2 + bx + c$. Each of the three points gives one equation in $a, b, c$. Solve the 3×3 system. See Section 7 for the technique.

Find the parabola with vertex $(-2, 3)$ passing through $(0, 7)$, in general form.

Start in vertex form: $y = a(x + 2)^2 + 3$.

Sub $(0, 7)$: $7 = a(2)^2 + 3 \Rightarrow 4a = 4 \Rightarrow a = 1$.

So $y = (x+2)^2 + 3 = x^2 + 4x + 7$.

Equivalent quadratics

If two quadratic functions are equal for all $x$, then when written in general form their corresponding coefficients must be identical. Expand, match $x^2$ coefficients, match $x$ coefficients, match constants — three equations.

Why this works Two polynomials can only agree at every single x-value if they are literally the same polynomial. There's no other way for it to happen.

Section 05 · Finding the Equation

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QUESTION 01

A parabola has $x$-intercepts at $1$ and $5$ with $a = 2$. Its equation in factored form is:

Why: Factored form is $y = a(x-\alpha)(x-\beta)$ with $\alpha, \beta$ the roots. So $y = 2(x-1)(x-5)$.

QUESTION 02

A parabola has vertex $(3, -1)$ and passes through $(4, 1)$. What is $a$ in $y = a(x-3)^2 - 1$?

Why: Sub $(4, 1)$: $1 = a(4-3)^2 - 1 = a - 1$, so $a = 2$.

QUESTION 03

If $f(x) = 3x^2 + 5x + 2$ and $g(x) = ax^2 + bx + c$ have the same graph for all $x$, what is $a + b + c$?

Why: Equivalent quadratics have identical coefficients. So $a = 3$, $b = 5$, $c = 2$, summing to $10$.

QUESTION 04

To find the equation of a parabola through three points, the easiest starting form is:

Why: Each point gives one equation in $a$, $b$, $c$ — three equations, three unknowns. General form turns it into a clean linear system.

QUESTION 05

If a parabola has vertex $(0, k)$, its axis of symmetry is:

Why: The axis is always the vertical line through the vertex. Vertex at $(0, k)$ ⇒ axis is $x = 0$, the $y$-axis.

Section 06

Intersection problems

Finding where a parabola meets a line is the same as finding the x-intercepts — just with a moved x-axis. The discriminant tells you whether they meet, touch, or miss entirely.

Solving $f(x) = k$ finds where $y = f(x)$ meets the horizontal line $y = k$. More generally, to find where $y = f(x)$ meets $y = g(x)$, set $f(x) = g(x)$ and solve.

Tangent, secant, or miss?

A horizontal line $y = c$ glides over the parabola $y = x^2$. Below the vertex it misses entirely. At the vertex it kisses the curve (tangent). Above, it slices through twice (secant).

Horizontal line y = 0.00

Discriminant 0.00

Status Tangent

Two key words A tangent meets a curve at exactly one point ($\Delta = 0$ after combining). A secant meets a curve at two distinct points ($\Delta > 0$).

Prove $y = 5x + 2$ is a tangent to $y = 2x^2 + x + 4$.

Set them equal: $2x^2 + x + 4 = 5x + 2$

$2x^2 - 4x + 2 = 0$

$\Delta = (-4)^2 - 4(2)(2) = 16 - 16 = 0$

$\Delta = 0$ means exactly one intersection point — so yes, it's a tangent. $\blacksquare$

Find $c$ if $y = 2x - 3$ is a secant to $y = -x^2 + 3x + c$.

Set equal: $-x^2 + 3x + c = 2x - 3$, so $-x^2 + x + (c+3) = 0$, or $x^2 - x - (c+3) = 0$.

Secant needs $\Delta > 0$: $(-1)^2 - 4(1)(-(c+3)) > 0$

$1 + 4c + 12 > 0 \Rightarrow 4c + 13 > 0 \Rightarrow c > -\tfrac{13}{4}$

Section 06 · Intersection Problems

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QUESTION 01

For $y = x^2 + 3$ to be tangent to $y = k$, what must $k$ equal?

Why: The vertex of $y = x^2 + 3$ is $(0, 3)$. A horizontal line tangent to the parabola must touch it at the vertex, so $k = 3$.

QUESTION 02

A line is a tangent to a parabola when, after setting them equal, the discriminant satisfies:

Why: Tangent ⇒ exactly one intersection point ⇒ one solution to the combined equation ⇒ $\Delta = 0$.

QUESTION 03

How many intersection points does a secant line have with a curve?

Why: "Secant" literally means a line that cuts a curve at two points. Hence $\Delta > 0$ once you equate them.

QUESTION 04

$y = x^2$ and $y = -1$ intersect at how many points?

Why: $x^2 = -1$ has no real solutions. The line $y = -1$ sits entirely below the parabola.

QUESTION 05

Find the points of intersection of $y = x^2$ and $y = x + 2$.

Why: $x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0$. So $x = 2$ giving $(2, 4)$, and $x = -1$ giving $(-1, 1)$.

Section 07

Simultaneous equations

Three techniques. Pick the right one and the problem is half-solved before you start computing.

Graphical

Best when sketches are quick

Graph both curves on the same axes. Read off the points of intersection. Limited by graph accuracy — only useful for nice integer solutions.

Substitution

Best when one is already y = ...

Substitute one equation into the other. Always the right move with a linear + quadratic pair, or two quadratics in $y = ...$ form.

Elimination

Best for two linears in $ax + by = c$

Add or subtract the equations (possibly after multiplying) so one variable cancels. Almost always quickest for linear systems.

Substitution with two quadratics

Find the intersection of $y = 2x^2 + 3x + 5$ and $y = x^2 - 4x - 7$.

Set equal: $2x^2 + 3x + 5 = x^2 - 4x - 7$

$x^2 + 7x + 12 = 0 \Rightarrow (x+3)(x+4) = 0$

$x = -3$: $y = (-3)^2 - 4(-3) - 7 = 9 + 12 - 7 = 14$ → $(-3, 14)$

$x = -4$: $y = 16 + 16 - 7 = 25$ → $(-4, 25)$

Elimination — the multiply-then-cancel trick

Solve $2x + 5y = 13$ and $3x - 4y = -15$.

To eliminate $x$, multiply the first by $3$ and the second by $2$:

$6x + 15y = 39$
$6x - 8y = -30$

Subtract: $23y = 69 \Rightarrow y = 3$

Sub into first: $2x + 15 = 13 \Rightarrow x = -1$

Finding a parabola through three points

Find the parabola passing through $(0, 2)$, $(3, 5)$ and $(4, 10)$.

Start with $y = ax^2 + bx + c$.

$(0, 2) \Rightarrow c = 2$

$(3, 5) \Rightarrow 9a + 3b + 2 = 5 \Rightarrow 9a + 3b = 3 \Rightarrow 3a + b = 1$

$(4, 10) \Rightarrow 16a + 4b + 2 = 10 \Rightarrow 16a + 4b = 8 \Rightarrow 4a + b = 2$

Subtract: $a = 1$, so $b = 1 - 3 = -2$.

$y = x^2 - 2x + 2$.

Section 07 · Simultaneous Equations

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QUESTION 01

Solve $y = 2x + 1$ and $y = -x + 4$.

Why: $2x + 1 = -x + 4 \Rightarrow 3x = 3 \Rightarrow x = 1$. Then $y = 2(1) + 1 = 3$.

QUESTION 02

Which method is most efficient for $3x + 2y = 12$ and $3x - 2y = 0$?

Why: Coefficients of $y$ are $+2$ and $-2$ — opposite. Adding eliminates $y$ instantly: $6x = 12 \Rightarrow x = 2$.

QUESTION 03

At most, how many points can two different parabolas intersect at?

Why: Setting them equal gives (at most) a quadratic equation, so up to $2$ solutions. The $x^2$ terms may even cancel, leaving a linear equation.

QUESTION 04

At most, how many points can a line and a parabola intersect at?

Why: Substituting gives a quadratic — so at most two real solutions (secant). Tangent gives one, missing gives zero.

QUESTION 05

To find a parabola through 3 points, how many equations and unknowns do you need?

Why: Three unknowns ($a$, $b$, $c$) need three independent equations to pin them down. Each point gives one equation.

Section 08

Quadratic inequalities

Don't try to solve these algebraically by "moving things across". Sketch the parabola, find where it sits above or below the x-axis, and read off the answer.

The graphical method

Move everything to one side so it's $\,(\text{quadratic}) \;\square\; 0\,$ where $\square$ is your inequality.
Find the roots (the x-intercepts).
Sketch the parabola roughly — you only really need the roots and concavity.
Shade the region where the parabola is on the correct side of the x-axis.
Read off the x-values of the shaded region.

Inequality direction For $f(x) > 0$, you want where the parabola is above the x-axis. For $f(x) < 0$, you want below. Use $\le$ / $\ge$ if the roots themselves are included (closed intervals); use $<$ / $>$ if not (open intervals).

Above and below the axis

Same concave-up parabola $y = (x-2)(x+1)$, with roots at $-1$ and $2$. The shaded segment on the x-axis flips between the two inequality directions — outside the roots when $f(x) > 0$, between the roots when $f(x) < 0$.

Currently solving f(x) > 0

Solution set x < −1 or x > 2

Region Outside the roots

Solve $x^2 + 3x + 2 \geq 0$.

Factor: $(x+1)(x+2) \geq 0$. Roots are $-1$ and $-2$. Concave up.

A concave-up parabola is at or above zero outside the roots — so on the wings.

Solution: $x \leq -2$ or $x \geq -1$.

Solve $x^2 - 25 < 0$.

$(x-5)(x+5) < 0$. Roots are $\pm 5$. Concave up.

Strictly below zero means between the roots.

Solution: $-5 < x < 5$.

Above the axis → outside the roots. Below the axis → between the roots. (Reversed if concave down.)

Solve $x^2 + 4x + 1 \leq 0$ (no nice factors — use the quadratic formula).

Roots: $x = \dfrac{-4 \pm \sqrt{16 - 4}}{2} = \dfrac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}$.

Concave up, so $\le 0$ is between the roots.

Solution: $-2 - \sqrt{3} \leq x \leq -2 + \sqrt{3}$.

Section 08 · Quadratic Inequalities

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QUESTION 01

Solve $(x-2)(x+3) > 0$.

Why: Concave up, roots $-3$ and $2$. Above the axis is outside the roots — so $x < -3$ or $x > 2$.

QUESTION 02

Solve $x^2 - 9 \leq 0$.

Why: $(x-3)(x+3) \le 0$. Concave up parabola is at or below the axis between the roots — including the roots themselves because of $\le$.

QUESTION 03

Solve $-x^2 + 4 \geq 0$.

Why: Rearrange: $x^2 \le 4$, so $-2 \le x \le 2$. (Or: concave-down parabola with roots $\pm 2$ — at or above zero between the roots.)

QUESTION 04

A concave-up parabola has roots at $-4$ and $1$. The solution to $f(x) < 0$ is:

Why: Below the axis on a concave-up parabola is between the roots — open interval because the inequality is strict.

QUESTION 05

Solve $x^2 + 1 > 0$.

Why: $\Delta = -4 < 0$ and $a = 1 > 0$ — the parabola is positive definite, sitting entirely above the axis. So every real $x$ works.

Section 09

Cubic functions

Power 3 instead of 2. Up to three x-intercepts, one inflection point, and a characteristic S-shape (or its mirror).

The three forms

Simple cubic

$y = kx^3$

Horizontal point of inflection at the origin. Slopes up-right if $k > 0$, down-right if $k < 0$. Bigger $|k|$ → steeper.

Factored form

$y = k(x-\alpha)(x-\beta)(x-\gamma)$

Three x-intercepts: $\alpha, \beta, \gamma$. The curve has two turning points and one inflection. Same sign rules for $k$.

Transformed form

$y = k(x-h)^3 + c$

The horizontal point of inflection sits at $(h, c)$. Same shape as $y = kx^3$, just shifted.

When two roots collide

Watch what happens as one root drifts toward another in $y = 0.3(x-a)(x-1.5)(x+1.5)$. When $a$ matches an existing root, two roots fuse into a double root — and the curve stops crossing the axis, instead just touching and bouncing.

Moving root a = 0.00

Number of distinct roots 3

Behaviour at moving root Crosses

Sketching simple cubics

Mark the inflection point (origin for $y = kx^3$, otherwise from the form).
Decide overall slope direction from sign of $k$.
Plot one extra point (often $x = 1$) to fix the steepness.

Sketching factored cubics

Mark all x-intercepts.
Find the y-intercept by substituting $x = 0$.
Use the sign of $k$: positive starts bottom-left, climbs up-right; negative does the opposite.
The curve crosses the x-axis at each simple root and "wiggles" between them.

Double roots — the "bounce"

Key idea A repeated factor like $(x-1)^2$ means the curve touches the x-axis at $x = 1$ without crossing it — it "bounces" off. This is the same idea as $\Delta = 0$ for parabolas.

Sketch $y = -(x-1)^2(x+1)$.

x-intercepts: $x = 1$ (double root → bounce) and $x = -1$ (simple root → cross).

y-intercept: $y = -(0-1)^2(0+1) = -1$.

Leading coefficient is $-1$ (negative), so the curve starts top-left, slopes overall downward to the right.

Shape: comes down from top-left, crosses at $x = -1$, dips down through $(0,-1)$, bounces off the x-axis at $x = 1$, then heads down to the bottom-right.

Sketching from general form

Factorise first — usually by taking out a common factor of $x$, then factorising what's left as a quadratic.

Sketch $y = x^3 + 7x^2 + 12x$.

Factor: $y = x(x^2 + 7x + 12) = x(x+3)(x+4)$

x-intercepts: $0$, $-3$, $-4$. All simple roots — curve crosses at each.

Positive leading coefficient → starts bottom-left, ends top-right, with two turning points between the intercepts.

Finding $k$ for $y = kx^3$

The curve $y = kx^3$ passes through $(-2, 2)$. Find $k$.

$2 = k(-2)^3 = -8k$, so $k = -\dfrac{1}{4}$.

Section 09 · Cubic Functions

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QUESTION 01

How many real $x$-intercepts can a cubic function have?

Why: A cubic runs from $-\infty$ to $+\infty$ (or the reverse), so it must cross the axis at least once. With a double root it can have 2; with three distinct roots, 3.

QUESTION 02

For $y = (x-1)^2(x-3)$, what happens at $x = 1$?

Why: $(x-1)^2$ is a repeated factor — that's a double root, which makes the curve touch and bounce off rather than cross.

QUESTION 03

For $y = kx^3$ with $k < 0$, the graph:

Why: Negative leading coefficient flips the standard $y = x^3$ shape vertically — so it goes from top-left to bottom-right.

QUESTION 04

The horizontal point of inflection of $y = 2(x - 3)^3 + 5$ is at:

Why: Transformed form $y = k(x-h)^3 + c$ has its HPI at $(h, c)$. Here $h = 3$, $c = 5$.

QUESTION 05

For $y = -(x+2)(x-1)(x-3)$, as $x \to +\infty$:

Why: Expanding gives a leading term of $-x^3$. Negative coefficient on $x^3$ means the graph drops on the right, so $y \to -\infty$.

The cheat sheet

Everything in one page

Quadratic

General form$y = ax^2 + bx + c$

Factored form$y = a(x - \alpha)(x - \beta)$

Vertex form$y = a(x - h)^2 + k$, vertex $(h, k)$

Quadratic formula$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Axis of symmetry$x = -\dfrac{b}{2a}$

Discriminant

Definition$\Delta = b^2 - 4ac$

Two distinct real roots$\Delta > 0$

Equal (double) root$\Delta = 0$

No real roots$\Delta < 0$

Rational roots$\Delta$ is a perfect square

Positive definite$\Delta < 0$ and $a > 0$

Negative definite$\Delta < 0$ and $a < 0$

Tangent condition$\Delta = 0$ (after equating curves)

Secant condition$\Delta > 0$ (after equating curves)

Inequalities

Above axis (concave up)Outside the roots

Below axis (concave up)Between the roots

Above axis (concave down)Between the roots

Below axis (concave down)Outside the roots

Cubic

General form$y = ax^3 + bx^2 + cx + d$

Factored form$y = k(x - \alpha)(x - \beta)(x - \gamma)$

Transformed form$y = k(x - h)^3 + c$

Double rootCurve "bounces" off the x-axis

$y = kx^3$, $k > 0$Slopes up left → right

$y = kx^3$, $k < 0$Slopes down left → right