Features of a Parabola (AI SL)
Before you can use a quadratic model, you usually need to read information straight off its graph: where it crosses the axes, where its turning point sits, and what shape it has. This page covers those key features - intercepts, axis of symmetry, vertex and roots - with worked examples and the mistakes that lose marks. It's part of the broader Quadratic Models topic.
20 questions on this sub-topic.
Reading off the key features
Covered under IB syllabus reference SL2.4: determine key features of graphs - maximum and minimum values, intercepts, symmetry, vertex, and zeros of functions or roots of equations - using graphing technology.
From \(y=ax^2+bx+c\)
y-intercept \(=c\), axis of symmetry \(x=-\dfrac{b}{2a}\)
The constant term \(c\) is the value of \(y\) when \(x=0\). The axis of symmetry passes through the vertex and splits the parabola into two mirror-image halves.
Factorised form
\(y = a(x-p)(x-q)\)
Here \(p\) and \(q\) are the roots (the \(x\)-intercepts). If you're given the roots and one other point, substitute the point in to find \(a\).
Need the full syllabus table and GDC walkthroughs? See Quadratic Models.
Worked examples
For \(y=x^{2}+2x-8\):
(a) Find the \(y\)-intercept.
(b) Find the two \(x\)-intercepts.
Worked solution
(a) \(y\)-intercept. Set \(x=0\): \(y=0^2+2(0)-8=-8\), so \((0,-8)\). A1
(b) \(x\)-intercepts.
Step 1 - Factorise.
Factorise \(x^2+2x-8=(x+4)(x-2)=0\) M1
\(x=-4\) A1 or \(x=2\). A1
A quadratic has roots \(x=-2\) and \(x=5\) and passes through \((0,-20).\) Find it in the form \(y=a(x-p)(x-q).\)
Worked solution
from the roots \(-2\) and \(5\): \(y=a(x+2)(x-5)\). M1
using \((0,-20)\): \(-20=a(0+2)(0-5)=a(2)(-5)=-10a\Rightarrow a=2.\) A1
A quadratic model is \(y=(x-3)(x-9).\) State the \(x\)-intercepts.
(a)(i) State the \(x\)-intercept with \(x<6\).
(a)(ii) State the \(x\)-intercept with \(x>6\).
(a)(iii) State the axis of symmetry.
Worked solution
(a)(i) \(x-3=0\Rightarrow x=3,\) A1
(a)(ii) \(x-9=0\Rightarrow x=9.\) A1
The vertex sits exactly halfway between the two roots, so the axis is their midpoint:
(a)(iii) \(x=\frac{3+9}{2}=6.\) A1The curve is symmetric about the vertical line \(x=6\).
Common mistakes
- Mixing up the sign of the roots. If the factorised form is \(y=a(x+4)(x-2)\), the roots are \(x=-4\) and \(x=2\) - not \(4\) and \(-2\). Flip the sign inside each bracket to get the actual root.
- Forgetting to solve for \(a\). Knowing the roots only fixes the shape of the parabola up to a vertical stretch - you still need a third point to pin down the value of \(a\) in \(y=a(x-p)(x-q)\).
- Confusing the axis of symmetry with the vertex. \(x=-\frac{b}{2a}\) gives you the \(x\)-coordinate of the vertex, not the vertex itself - substitute it back into the equation to get the \(y\)-coordinate too.
Ready to practise properly?
20 features-of-a-parabola questions, marked instantly like the real exam.
Quick answers
How do you find the y-intercept of a quadratic?
Set \(x=0\) in \(y=ax^2+bx+c\). Every \(x\)-term drops out, leaving \(y=c\) - the \(y\)-intercept is always the constant term.
How do you find a quadratic from its roots?
Write it in factorised form \(y=a(x-p)(x-q)\) using the two roots \(p\) and \(q\), then substitute a known point into the equation to solve for \(a\).