Graphing ax^2+bx+c
The graph of \(y = ax^2 + bx + c\) is a parabola, concave up when \(a \gt 0\) and concave down when \(a \lt 0\), with its axis of symmetry at \(x = -\dfrac{b}{2a}\). This NSW Year 9 Advanced maths subtopic sketches parabolas from their key features (NESA outcome MA5‑NLI‑P‑01).
You will find \(x\)- and \(y\)-intercepts by factorising, find the vertex and axis of symmetry, and sketch parabolas such as \(y = 2x^2 - 4x - 6\). You will describe how \(y = (x - 3)^2 + 2\) is moved from \(y = x^2\), find an equation from intercepts, find where a line meets a parabola, and find the values of \(k\) that give two, one or no \(x\)-intercepts. It is part of Year 9 Advanced Coordinate Geometry and Graphs.
Every question with a fully worked solution.
- Visual introduction to parabolas Watch
Theory
The graph of \(y = ax^2 + bx + c\) is a parabola. To sketch it, read the concavity from \(a\), find the intercepts, then the axis of symmetry and vertex. This Year 9 Advanced maths page (NSW Stage 5, Non-linear relationships) is part of the path towards Advanced and Extension maths in Years 11 and 12.
A parabola is the graph of a quadratic, \(y = ax^2 + bx + c\) with \(a \ne 0\). It is symmetric about a vertical line called the axis of symmetry.
The vertex (or turning point) is where the axis meets the curve. It is the lowest point (a minimum) when the parabola is concave up, and the highest point (a maximum) when it is concave down.
The sign of \(a\) alone decides the concavity: \(a > 0\) concave up, \(a < 0\) concave down. The constant \(c\) is the \(y\)-intercept, because \(x = 0\) leaves \(y = c\).
The \(x\)-intercepts are the solutions of \(ax^2 + bx + c = 0\). A parabola can have \(2\), \(1\) or \(0\) of them. When the equation factorises as \(y = a(x - p)(x - q)\), the \(x\)-intercepts are \(p\) and \(q\).
In turning-point form, \(y = a(x - h)^2 + k\), the vertex is \((h, k)\): the graph of \(y = ax^2\) moved \(h\) units across and \(k\) units up.
The axis of symmetry of \(y = ax^2 + bx + c\):
If the \(x\)-intercepts are \(p\) and \(q\), the axis is halfway between them:
Turning-point form, vertex \((h, k)\):
How to sketch y = ax² + bx + c
- Concavity: \(a > 0\) concave up, \(a < 0\) concave down.
- \(y\)-intercept: put \(x = 0\); it is \(c\).
- \(x\)-intercepts: put \(y = 0\). Take out any common factor (including \(-1\)), then factorise and solve.
- Axis: \(x = -\dfrac{b}{2a}\), or the midpoint of the \(x\)-intercepts.
- Vertex: substitute the axis value into the equation to get \(y\).
- Draw a smooth, symmetric curve through the points. Label every intercept and the vertex. Use symmetry for an extra point: the mirror image of the \(y\)-intercept.
Working backwards and with unknowns
- From the intercepts: write \(y = a(x - p)(x - q)\), substitute one more point to find \(a\), then expand.
- Finding \(b\) and \(c\): when \(a = 1\) and the \(x\)-intercepts are known, expand \((x - p)(x - q)\) and compare coefficients.
- Unknown constant: find the vertex in terms of the constant, then use 'below, on or above the \(x\)-axis' to count the intercepts.
- Areas and lengths: once the intercepts and vertex are known, a triangle on the \(x\)-axis has base = distance between the intercepts and height = the vertex's \(y\)-value (as a length).
Find the \(y\)-intercept, factorise for the \(x\)-intercepts, then the vertex sits halfway between them.
| \(x = 0:\ y\) | \(=\) | \(-15\) |
| \(x^2 + 2x - 15\) | \(=\) | \(0\) |
| \((x + 5)(x - 3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-5 \text{ or } x = 3\) |
| \(\text{axis: } x\) | \(=\) | \(\dfrac{-5 + 3}{2} = -1\) |
| \(y\) | \(=\) | \((-1)^2 + 2(-1) - 15\) |
| \(\) | \(=\) | \(\mathbf{-16}\) |
\(a = 1 > 0\), so the parabola is concave up and the vertex \((-1, -16)\) is a minimum.
Take out the common factor \(-2\) first, so the bracket is monic and easy to factorise.
| \(x = 0:\ y\) | \(=\) | \(10\) |
| \(-2x^2 - 8x + 10\) | \(=\) | \(0\) |
| \(-2(x^2 + 4x - 5)\) | \(=\) | \(0\) |
| \(-2(x + 5)(x - 1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-5 \text{ or } x = 1\) |
Use the formula for the axis as a check on the midpoint.
| \(x\) | \(=\) | \(-\dfrac{b}{2a} = -\dfrac{-8}{2 \times (-2)} = -2\) |
| \(y\) | \(=\) | \(-2(-2)^2 - 8(-2) + 10\) |
| \(\) | \(=\) | \(-8 + 16 + 10 = \mathbf{18}\) |
\(a = -2 < 0\): concave down, so \((-2, 18)\) is a maximum turning point.
An \(x\)-intercept at \(-3\) gives the factor \((x + 3)\); at \(1\) it gives \((x - 1)\). The value of \(a\) is still unknown.
| \(y\) | \(=\) | \(a(x + 3)(x - 1)\) |
| \(8\) | \(=\) | \(a(-1 + 3)(-1 - 1)\) |
| \(8\) | \(=\) | \(-4a\) |
| \(a\) | \(=\) | \(-2\) |
Expand:
| \(y\) | \(=\) | \(-2(x^2 + 2x - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{-2x^2 - 4x + 6}\) |
Check the \(y\)-intercept: \(-2(3)(-1) = 6\) \(\checkmark\)
(i) Describe how \(y = x^2\) is moved to give \(y = (x - 1)^2 - 9\), and state the vertex.
(ii) Find the intercepts.
(iii) For which values of \(c\) does \(y = (x - 1)^2 + c\) have no \(x\)-intercepts?
(i) The bracket is \(0\) when \(x = 1\): a shift of \(1\) unit right. The \(-9\) shifts it \(9\) units down. The vertex moves from \((0, 0)\) to \((1, -9)\).
(ii) Intercepts:
| \(x = 0:\ y\) | \(=\) | \((0 - 1)^2 - 9 = \mathbf{-8}\) |
| \((x - 1)^2 - 9\) | \(=\) | \(0\) |
| \((x - 1)^2\) | \(=\) | \(9\) |
| \(x - 1\) | \(=\) | \(\pm 3\) |
| \(x\) | \(=\) | \(\mathbf{4} \text{ or } \mathbf{-2}\) |
(iii) The vertex is \((1, c)\) and the parabola is concave up. It misses the \(x\)-axis when the vertex is above it: \(c > 0\).
Common pitfalls
Frequently asked questions
How do you find the vertex of a parabola?
Find the axis of symmetry with \(x = -\dfrac{b}{2a}\), or take the midpoint of the \(x\)-intercepts. Then substitute that \(x\)-value into the equation to find the \(y\)-value of the vertex.
How do you know if a parabola is concave up or down?
Look at \(a\), the coefficient of \(x^2\). If \(a > 0\) the parabola is concave up and has a minimum. If \(a < 0\) it is concave down and has a maximum. The values of \(b\) and \(c\) do not change the concavity.
How do you find the x-intercepts of a parabola?
Put \(y = 0\) and solve the quadratic equation, usually by factorising. Each solution is an \(x\)-intercept. If the equation has no solution, the parabola does not cross the \(x\)-axis.
Can a parabola have no x-intercepts?
Yes. A concave-up parabola whose vertex is above the \(x\)-axis, or a concave-down parabola whose vertex is below it, never meets the \(x\)-axis. For example \(y = x^2 + 4\) has no \(x\)-intercepts.
How do you find the equation of a parabola from its graph?
Read the \(x\)-intercepts \(p\) and \(q\) and write \(y = a(x - p)(x - q)\). Substitute another point on the graph, such as the \(y\)-intercept, to find \(a\). Then expand if the form \(ax^2 + bx + c\) is needed.
What does y = (x - 4) squared + 1 mean?
It is the parabola \(y = x^2\) moved \(4\) units right and \(1\) unit up. Its vertex is \((4, 1)\) and its axis of symmetry is \(x = 4\).