Graphing Quadratics
Learn to graph quadratics for Queensland Year 11 Mathematical Methods (QCAA). Every quadratic graphs as a parabola whose key features follow from its factor form and turning-point form.
You will learn to find the direction of opening, the turning point, the axis of symmetry and the intercepts, keep surd intercepts exact, and work back to a parabola's equation from its features — groundwork for the completing-the-square and modelling work ahead.
Every question with a fully worked solution.
- Graphing Quadratics - Video - 3 forms of the quadratic equation Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a quadratic function graphs as a parabola. This page shows how to read its key features — direction of opening, vertex (turning point), axis of symmetry and intercepts — from factor form \(y=a(x-x_1)(x-x_2)\) and vertex form \(y=a(x-h)^2+k\), and how to find a parabola's equation from its features.
A parabola is the graph of \(y=ax^2+bx+c\). It opens upward when \(a>0\) and downward when \(a<0\); a bigger \(|a|\) makes it narrower.
The vertex (turning point) is the lowest point if the parabola opens up, or the highest if it opens down. The axis of symmetry is the vertical line \(x=h\) through the vertex; the parabola is a mirror image in it.
The \(x\)-intercepts (zeros) are found by setting \(y=0\); the \(y\)-intercept by setting \(x=0\). In factor form \(y=a(x-x_1)(x-x_2)\) the zeros are \(x_1\) and \(x_2\); in vertex form \(y=a(x-h)^2+k\) the vertex is \((h,k)\) directly.
Reading features from the two useful forms:
How to sketch a parabola
- Opening: read the sign of \(a\) — up if \(a>0\), down if \(a<0\).
- Intercepts: set \(y=0\) for the \(x\)-intercepts and \(x=0\) for the \(y\)-intercept.
- Vertex and axis: find the axis of symmetry, substitute it back for the vertex \(y\)-value, then plot the points and draw a smooth curve.
Opening — expanding gives \(y=x^2-6x+5\), so \(a=1>0\): the parabola opens upward.
\(x\)-intercepts — set each factor to zero:
| \(x-1=0\) | \(=\) | \(x=1\) |
| \(x-5=0\) | \(=\) | \(x=5\) |
Axis of symmetry — midway between the zeros:
| \(x\) | \(=\) | \(\dfrac{1+5}{2}=3\) |
Vertex \(y\)-value — substitute \(x=3\):
| \(y\) | \(=\) | \((3-1)(3-5)=(2)(-2)=-4\) |
\(y\)-intercept — substitute \(x=0\):
| \(y\) | \(=\) | \((0-1)(0-5)=5\) |
Opens up; zeros \((1,0),(5,0)\); vertex \((3,-4)\); axis \(x=3\); \(y\)-intercept \((0,5)\).
Vertex — read \((h,k)\) straight from the form \(y=(x-2)^2-9\):
| \((h,k)\) | \(=\) | \((2,\,-9)\) |
\(x\)-intercepts — set \(y=0\) and take square roots:
| \((x-2)^2-9\) | \(=\) | \(0\) |
| \((x-2)^2\) | \(=\) | \(9\) |
| \(x-2\) | \(=\) | \(\pm3\) |
| \(x\) | \(=\) | \(5\ \text{or}\ -1\) |
\(y\)-intercept — substitute \(x=0\):
| \(y\) | \(=\) | \((0-2)^2-9=4-9=-5\) |
Opens up; vertex \((2,-9)\); axis \(x=2\); zeros \((-1,0),(5,0)\); \(y\)-intercept \((0,-5)\).
Vertex — complete the square to read \((h,k)\):
| \(y\) | \(=\) | \((x-2)^2-4+1\) |
| \(=\) | \((x-2)^2-3\) |
So the vertex is \((2,\,-3)\) and the axis is \(x=2\).
\(x\)-intercepts — set \(y=0\) and take square roots (leave as surds):
| \((x-2)^2-3\) | \(=\) | \(0\) |
| \((x-2)^2\) | \(=\) | \(3\) |
| \(x-2\) | \(=\) | \(\pm\sqrt{3}\) |
| \(x\) | \(=\) | \(2\pm\sqrt{3}\) |
Vertex \((2,-3)\); exact zeros \(x=2-\sqrt{3}\) and \(x=2+\sqrt{3}\).
Use vertex form with \((h,k)=(2,-3)\):
| \(y\) | \(=\) | \(a(x-2)^2-3\) |
Substitute the known point \((0,5)\) to find \(a\):
| \(5\) | \(=\) | \(a(0-2)^2-3\) |
| \(5\) | \(=\) | \(4a-3\) |
| \(4a\) | \(=\) | \(8\) |
| \(a\) | \(=\) | \(2\) |
Write the equation (and expand if required):
| \(y\) | \(=\) | \(2(x-2)^2-3\) |
| \(=\) | \(2x^2-8x+5\) |
\(y=2(x-2)^2-3=2x^2-8x+5\).
Common pitfalls
Frequently asked questions
How do I find the axis of symmetry from factor form?
Average the two \(x\)-intercepts: \(x=\dfrac{x_1+x_2}{2}\). The axis passes through the vertex.
How do I read the vertex from \(y=a(x-h)^2+k\)?
The vertex is \((h,k)\). The \(x\)-coordinate is the value making the bracket zero, and \(k\) is the constant added on.
Which way does the parabola open?
Upward if \(a>0\) (vertex is a minimum) and downward if \(a<0\) (vertex is a maximum).
How do I find the y-intercept?
Substitute \(x=0\) into the equation. In \(y=ax^2+bx+c\) that is simply the constant \(c\).
What if the x-intercepts are surds?
Set \(y=0\) and solve; if the result is irrational, leave it exact, for example \(x=2\pm\sqrt{3}\).
How do I find a parabola's equation from its turning point?
Write \(y=a(x-h)^2+k\) with the turning point \((h,k)\), then substitute one more known point to solve for \(a\).