Graphing Quadratics In Polynomial Form
Learn to graph quadratics in polynomial form for Queensland Year 11 Mathematical Methods (QCAA). This means sketching a parabola from a squared term, a linear term and a constant.
You will learn to find the opening direction, read the y-intercept from the constant, locate the axis of symmetry and turning point, and find the x-intercepts by factorising or completing the square — the standard sketch examiners expect for any parabola.
Every question with a fully worked solution.
- Graphing Quadratics In Polynomial Form - Video - Graphing Quadratics in polynomial form Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a quadratic in polynomial form \(y=ax^2+bx+c\) can be graphed straight from its coefficients: the \(y\)-intercept is \(c\), the axis of symmetry is \(x=-\dfrac{b}{2a}\), the turning point comes from substituting that \(x\), and the \(x\)-intercepts come from factorising. This page combines those features into a full sketch.
Polynomial form (also called general form) is \(y=ax^2+bx+c\). The sign of \(a\) gives the opening direction — up if \(a>0\), down if \(a<0\) — and the constant \(c\) is the \(y\)-intercept, since \(x=0\) gives \(y=c\).
The axis of symmetry is the vertical line \(x=-\dfrac{b}{2a}\). Substituting this \(x\)-value back into the equation gives the \(y\)-coordinate of the turning point (vertex).
The \(x\)-intercepts (zeros) are found by setting \(y=0\) and factorising; if the quadratic will not factorise over the rationals, complete the square to get exact surd answers.
Reading a parabola from \(y=ax^2+bx+c\):
How to sketch \(y=ax^2+bx+c\)
- Opening and \(y\)-intercept: read the sign of \(a\) and plot \((0,c)\).
- Axis and vertex: compute \(x=-\dfrac{b}{2a}\), then substitute it back for the turning point's \(y\)-value.
- \(x\)-intercepts: set \(y=0\) and factorise (or complete the square for surds), then draw a smooth parabola through all the points.
Opening and \(y\)-intercept — \(a=1>0\) opens up; \(c=8\) gives \(y\)-intercept \((0,8)\).
Axis of symmetry — use \(x=-\dfrac{b}{2a}\) with \(a=1,\ b=-6\):
| \(x\) | \(=\) | \(-\dfrac{-6}{2(1)}=3\) |
Turning point \(y\)-value — substitute \(x=3\):
| \(y\) | \(=\) | \((3)^2-6(3)+8\) |
| \(=\) | \(9-18+8=-1\) |
\(x\)-intercepts — set \(y=0\) and factorise:
| \(x^2-6x+8\) | \(=\) | \(0\) |
| \((x-2)(x-4)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(2\ \text{or}\ 4\) |
Opens up; \(y\)-intercept \((0,8)\); axis \(x=3\); vertex \((3,-1)\); zeros \((2,0),(4,0)\).
Axis of symmetry — \(a=1,\ b=4\):
| \(x\) | \(=\) | \(-\dfrac{4}{2(1)}=-2\) |
Turning point \(y\)-value — substitute \(x=-2\):
| \(y\) | \(=\) | \((-2)^2+4(-2)-5\) |
| \(=\) | \(4-8-5=-9\) |
\(x\)-intercepts — set \(y=0\) and factorise:
| \((x+5)(x-1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-5\ \text{or}\ 1\) |
\(y\)-intercept: \(c=-5\), so \((0,-5)\).
Vertex \((-2,-9)\); axis \(x=-2\); zeros \((-5,0),(1,0)\); \(y\)-intercept \((0,-5)\).
Axis of symmetry — \(a=1,\ b=-6\):
| \(x\) | \(=\) | \(-\dfrac{-6}{2(1)}=3\) |
Vertex \(y\)-value — substitute \(x=3\):
| \(y\) | \(=\) | \((3)^2-6(3)+7\) |
| \(=\) | \(9-18+7=-2\) |
\(x\)-intercepts — it will not factorise, so complete the square and take roots:
| \((x-3)^2-2\) | \(=\) | \(0\) |
| \((x-3)^2\) | \(=\) | \(2\) |
| \(x-3\) | \(=\) | \(\pm\sqrt{2}\) |
| \(x\) | \(=\) | \(3\pm\sqrt{2}\) |
Vertex \((3,-2)\); exact zeros \(x=3-\sqrt{2}\) and \(x=3+\sqrt{2}\).
Opening and \(y\)-intercept — \(a=-1<0\) opens down; \(c=3\) gives \((0,3)\).
Axis of symmetry — \(a=-1,\ b=2\):
| \(x\) | \(=\) | \(-\dfrac{2}{2(-1)}=1\) |
Turning point \(y\)-value — substitute \(x=1\):
| \(y\) | \(=\) | \(-(1)^2+2(1)+3\) |
| \(=\) | \(-1+2+3=4\) |
\(x\)-intercepts — set \(y=0\); factor out \(-1\) then factorise:
| \(-x^2+2x+3\) | \(=\) | \(0\) |
| \(-(x^2-2x-3)\) | \(=\) | \(0\) |
| \(-(x-3)(x+1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(3\ \text{or}\ -1\) |
Opens down; vertex \((1,4)\) is a maximum; zeros \((-1,0),(3,0)\); \(y\)-intercept \((0,3)\).
Common pitfalls
Frequently asked questions
How do I find the axis of symmetry from \(y=ax^2+bx+c\)?
Use \(x=-\dfrac{b}{2a}\). This vertical line passes through the turning point.
How do I get the turning point?
Find the axis \(x=-\dfrac{b}{2a}\), then substitute that \(x\) back into the equation to get the \(y\)-coordinate.
What is the y-intercept of a quadratic in polynomial form?
It is the constant term \(c\), because putting \(x=0\) gives \(y=c\). The point is \((0,c)\).
How do I find the x-intercepts?
Set \(y=0\) and factorise \(ax^2+bx+c=0\). If it will not factorise over the rationals, complete the square for exact surd answers.
Which way does the parabola open?
Upward if \(a>0\) and downward if \(a<0\); the size of \(|a|\) controls how narrow it is.
Do I need to factorise to find the turning point?
No. The turning point comes from \(x=-\dfrac{b}{2a}\) and substitution; factorising is only needed for the \(x\)-intercepts.