Every graph you have met so far has been a straight line, and straight lines can only ever go up or down at a steady rate.
A huge amount of the real world is not steady: a ball thrown in the air speeds up, slows, stops, and falls back, and a company's profit rises then drops as the price changes.
The quadratic function is the simplest possible way to bend a straight relationship into a curve with a single turning point, which is exactly what you need to model "rising then falling" or "falling then rising".
That single turning point is usually the whole reason we care: it is the highest a ball reaches, the price that gives maximum profit, or the least material needed to build something.
This is why quadratics appear everywhere in the course and in the exam: master the parabola and you can find a "best" value long before you meet calculus.
Analogy
Picture the path of a basketball shot: it climbs, curves over at the top, and drops into the net.
That arc is a parabola, the graph of a quadratic.
The peak of the arc is the vertex, and the fact that the way up mirrors the way down is the axis of symmetry.
A Quadratic Is A Polynomial Of Degree Two
A quadratic function is any function you can write in the form below, where $a$, $b$, and $c$ are real numbers: $$ f(x) = ax^2 + bx + c, \quad a \neq 0 $$
The highest power of $x$ is $2$, which is what the word "quadratic" refers to (from the Latin quadratus, meaning square).
The condition $a \neq 0$ is essential:
if $a$ were $0$ the $x^2$ term would vanish and you would be left with $f(x) = bx + c$, which is just a straight line.
So $a \neq 0$ is precisely the condition that keeps the graph curved.
The graph of every quadratic is a parabola: a smooth, symmetric, U-shaped curve.
If $a > 0$ the parabola opens upward and has a lowest point, and if $a < 0$ it opens downward and has a highest point.
Note
"Root", "zero", and "$x$-intercept" all mean the same thing for a quadratic: a value of $x$ where $f(x) = 0$.
A parabola can cross the $x$-axis twice, touch it once, or miss it entirely, so a quadratic can have two, one, or no real roots.
The Anatomy Of A Parabola
Before choosing a method, it helps to name the four features every exam question asks about.
The vertex is the single turning point, either the minimum or the maximum.
The axis of symmetry is the vertical line through the vertex that reflects one half of the curve onto the other.
The y-intercept is where the curve crosses the $y$-axis.
The x-intercepts are where the curve crosses the $x$-axis, if it does.
Standard Form: What a, b, And c Tell You At A Glance
The form $f(x) = ax^2 + bx + c$ is called standard form (or general form), and each coefficient controls a different feature.
Coefficient a
The sign of $a$ sets the direction: $a>0$ opens upward (concave up), $a<0$ opens downward (concave down).
The size of $a$ sets the width: a larger $|a|$ gives a narrower, steeper parabola.
Coefficient c
Substituting $x=0$ gives $f(0) = a(0)^2 + b(0) + c = c$, so the $y$-intercept is always the point $(0,\,c)$.
This is the fastest reading of all: $c$ is simply where the curve meets the $y$-axis.
Coefficient b
On its own $b$ has no clean meaning, but together with $a$ it fixes the axis of symmetry, which we derive next.
Example question
For $f(x) = 2x^2 - 4x + 3$, state the direction the parabola opens and write down its $y$-intercept.
Solution
The parabola opens upward because $a = 2 > 0$.
The $y$-intercept is $(0,\,3)$ because $c = 3$.
Deriving The Axis Of Symmetry
The axis of symmetry is the mirror line of the parabola, and its equation is given in the formula booklet (Topic 2: Functions): $$ x = -\frac{b}{2a} $$
It is worth seeing where it comes from, because the same algebra hands you the vertex for free.
Group the two terms that contain $x$ and factor out $a$: $$ f(x) = a\left(x^2 + \tfrac{b}{a}x\right) + c $$
Inside the bracket, complete the square by adding and subtracting the square of half the coefficient of $x$, that is $\left(\tfrac{b}{2a}\right)^2$: $$ f(x) = a\left(x^2 + \tfrac{b}{a}x + \tfrac{b^2}{4a^2} - \tfrac{b^2}{4a^2}\right) + c $$
The first three terms inside are a perfect square, so fold them up: $$ f(x) = a\left(x + \tfrac{b}{2a}\right)^2 - \tfrac{b^2}{4a} + c $$
A squared bracket is never negative, so the whole expression is smallest (or largest, if $a<0$) exactly when the bracket equals zero.
That happens when $x + \tfrac{b}{2a} = 0$, which rearranges to $x = -\tfrac{b}{2a}$.
Tip
The turning point sits on this line, so $x = -\frac{b}{2a}$ is both the axis of symmetry and the $x$-coordinate of the vertex.
Exam technique
Because this formula is in the booklet, exam marks come from selecting and applying it, not from reciting it.
A common Paper 1 task gives you the axis of symmetry and one other fact and asks you to work back to $a$, $b$, or $c$.
Reading The Vertex From The Axis
Once you have the $x$-coordinate of the vertex from $x = -\tfrac{b}{2a}$, substitute it back into $f(x)$ to get the $y$-coordinate.
The vertex is therefore the point $\left(-\tfrac{b}{2a},\; f\!\left(-\tfrac{b}{2a}\right)\right)$.
Example question
Find the coordinates of the vertex of $f(x) = 2x^2 - 4x + 3$.
Solution
Axis of symmetry: $x = -\dfrac{b}{2a} = -\dfrac{-4}{2(2)} = 1$.
Substitute $x=1$: $f(1) = 2(1)^2 - 4(1) + 3 = 1$.
Vertex: $(1,\,1)$, a minimum since $a>0$.
Vertex Form Makes The Turning Point Visible
The vertex form of a quadratic is written with the turning point built directly into the equation: $$ f(x) = a(x-h)^2 + k $$
The vertex is the point $(h,\,k)$, and the axis of symmetry is the line $x = h$.
You can read the turning point straight off, and the reason is the squared bracket:
$(x-h)^2 \geq 0$ for every $x$, and it equals $0$ only when $x = h$.
So when $a>0$ the smallest value of $f$ is $k$, reached at $x=h$; when $a<0$ that same point is the maximum.
Comparing with the completed-square result from the derivation above, $h = -\tfrac{b}{2a}$ and $k = f(h)$, so vertex form is simply standard form with the turning point already exposed.
Example question
The function $f(x) = 3(x-2)^2 - 5$ is in vertex form. State its vertex, its axis of symmetry, and whether it has a maximum or minimum.
Solution
Vertex: $(2,\,-5)$.
Axis of symmetry: $x = 2$.
Since $a = 3 > 0$, the vertex is a minimum.
Completing The Square: Standard Form Into Vertex Form
To convert a specific quadratic from standard to vertex form, follow the same steps as the general derivation, but with numbers.
Take $f(x) = x^2 - 6x + 5$ as a worked example.
Look at the coefficient of $x$, which is $-6$, and halve it to get $-3$.
Add and subtract the square of that number, $(-3)^2 = 9$: $$ x^2 - 6x + 5 = \left(x^2 - 6x + 9\right) - 9 + 5 $$
The bracket is now a perfect square, $(x-3)^2$, so rewrite it: $$ f(x) = (x-3)^2 - 4 $$
Read off the vertex directly: $(3,\,-4)$, a minimum because $a = 1 > 0$.
When $a \neq 1$, factor $a$ out of the $x^2$ and $x$ terms first, exactly as in the general derivation, then complete the square inside the bracket.
Common Mistake
Do not forget to subtract the number you added: adding $9$ without subtracting it changes the function.
When you factor $a$ out first, remember to multiply back through when you take the constant outside the bracket.
The sign flips: $(x-3)^2$ gives $h = +3$, not $-3$, because the form is $(x-h)^2$.
Example question
Express $f(x) = 2x^2 + 8x + 3$ in the form $a(x-h)^2 + k$ and hence state the minimum value of $f$.
Solution
Factor $2$ from the first two terms: $$f(x) = 2(x^2 + 4x) + 3$$
Half of $4$ is $2$, and $2^2 = 4$, so add and subtract $4$ inside: $$f(x) = 2(x^2 + 4x + 4 - 4) + 3$$
Fold the perfect square and take the extra term out (remembering the factor $2$): $$f(x) = 2(x+2)^2 - 8 + 3$$
Simplify: $$f(x) = 2(x+2)^2 - 5$$
The minimum value is $-5$, at $x=-2$.
Factored Form Reveals The x-Intercepts
The factored form (or intercept form) writes the quadratic as a product: $$ f(x) = a(x-p)(x-q) $$
The $x$-intercepts are $(p,\,0)$ and $(q,\,0)$, and here is why:
a product equals zero only when one of its factors is zero, so $f(x)=0$ forces $x=p$ or $x=q$.
This is the zero product property, and it is the whole reason factored form is so quick for finding roots.
The axis of symmetry sits exactly halfway between the two roots, at $x = \dfrac{p+q}{2}$, because the parabola is symmetric about its vertex.
Note
Not every quadratic factors nicely over the real numbers; some have no real roots and so cannot be written in this form.
Discriminant (covered in SL 2.7) tells you in advance how many real roots exist.
Example question
A quadratic has $x$-intercepts at $x = -1$ and $x = 4$ and passes through $(0, -8)$. Find $f(x)$ in the form $ax^2+bx+c$.
Solution
Start from factored form: $f(x) = a(x+1)(x-4)$.
Use the point $(0,-8)$: $-8 = a(1)(-4) = -4a$, so $a = 2$.
Expand: $f(x) = 2(x+1)(x-4) = 2(x^2 - 3x - 4)$.
Final answer: $f(x) = 2x^2 - 6x - 8$.
Choosing The Right Form For The Job
All three forms describe the same parabola, so the skill is picking the one that exposes what a question asks for.
Form
Equation
Reads off instantly
Standard
ax² + bx + c
y-intercept (0, c)
Factored
a(x - p)(x - q)
x-intercepts p and q
Vertex
a(x - h)² + k
vertex (h, k)
How To Move Between Them
To go from factored or vertex form to standard form, expand the brackets and collect like terms.
To go from standard to vertex form, complete the square, as shown above.
To go from standard to factored form, factorise, or solve $f(x)=0$ for the roots $p$ and $q$.
Tip
If a question asks for the maximum or minimum, aim for vertex form.
If it asks where the graph crosses the $x$-axis, aim for factored form.
If it asks where the graph crosses the $y$-axis, standard form already gives it as $c$.
Example question
The quadratic $f(x) = x^2 - 6x + 5$ was written in vertex form as $f(x) = (x-3)^2 - 4$ above. Write it in factored form and state all three intercepts.
Solution
Factorise the original: $$x^2 - 6x + 5 = (x-1)(x-5)$$
Factored form: $$f(x) = (x-1)(x-5)$$
$x$-intercepts: $(1,0)$ and $(5,0)$; $y$-intercept: $(0,5)$; vertex: $(3,-4)$.
Vertex Form Is The Base Parabola, Moved
Vertex form is the clearest example of transformations of a graph, which you study fully in SL 2.11.
Start from the base parabola $y = x^2$, whose vertex sits at the origin.
The $-h$ inside the bracket shifts it horizontally, moving the vertex to $x = h$.
The $+k$ outside shifts it vertically, moving the vertex to height $k$.
The factor $a$ stretches it vertically, and a negative $a$ also reflects it in the $x$-axis so it opens downward.
Tip
Reading $a$, $h$, and $k$ off vertex form tells you exactly how the base parabola has been moved and stretched to make your curve.
Example
The graph of $f(x) = -2(x-1)^2 + 3$ is the base parabola $y=x^2$ stretched by factor $2$, reflected to open downward, then moved to the vertex $(1, 3)$.
Its maximum value is $3$, reached at $x = 1$.
Active recall
Why must $a \neq 0$ for a function to be quadratic, and what shape is its graph?
State the formula for the axis of symmetry and explain how it also gives the $x$-coordinate of the vertex.
Which of the three forms lets you read off the vertex, and which lets you read off the $x$-intercepts?
Describe the steps to convert $f(x) = x^2 - 6x + 5$ into vertex form by completing the square.
Starting from $y = x^2$, what four changes does $f(x) = a(x-h)^2 + k$ describe?
