Exploring Parabolas
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- Exploring Parabolas - Video - Visual introduction to parabolas Watch
Theory
The graph of a quadratic \(y = ax^2 + bx + c\) is a parabola. This Year 10A maths page shows how to draw one from a table of values and read its key features: the turning point, the axis of symmetry, the concavity, the intercepts and the range. It derives the turning point \(\left(-\dfrac{b}{2a},\ c - \dfrac{b^2}{4a}\right)\), uses horizontal lines to count solutions, and finds unknown coefficients.
A parabola is the graph of a quadratic \(y = ax^2 + bx + c\) with \(a \ne 0\). If \(a \gt 0\) it is concave up (opens upwards); if \(a \lt 0\) it is concave down.
The \(x\)-intercepts are where \(y = 0\); a parabola can have two, one or none. The \(y\)-intercept is where \(x = 0\), which is always \(y = c\).
The axis of symmetry is the vertical line that folds the parabola onto itself. The turning point (or vertex) lies on it: a minimum if the parabola is concave up, a maximum if it is concave down.
The range is the set of \(y\)-values the graph reaches. A concave up parabola with turning point \((h, k)\) has range \(y \ge k\); a concave down one has range \(y \le k\).
In \(y = ax^2\), the size of \(a\) sets the width: the larger \(a\) is, the narrower (steeper) the parabola; a value between \(0\) and \(1\) makes it wider than \(y = x^2\).
Axis of symmetry of \(y = ax^2 + bx + c\), and the axis from two \(x\)-intercepts \(p\) and \(q\):
Completing the square when \(a = 1\) gives the turning point directly:
How to sketch a parabola
- Concavity: read the sign of \(a\).
- \(y\)-intercept: \(y = c\).
- \(x\)-intercepts: let \(y = 0\) and solve, by factorising if you can.
- Axis and turning point: \(x = -\dfrac{b}{2a}\) or the midpoint of the \(x\)-intercepts; substitute to get the \(y\)-value.
- Sketch a smooth curve, symmetric about the axis, through the labelled points. Use symmetry for an extra point: the mirror image of the \(y\)-intercept.
Reading a graph or a table
- The lowest (or highest) point is the turning point; the axis is \(x =\) its \(x\)-coordinate.
- In a table, look for the \(y\)-values that repeat: the axis is halfway between their \(x\)-values. The \(x\)-intercepts are where \(y = 0\).
- Use the turning point to state the range.
Finding unknown coefficients
- Three points: \((0, c)\) gives \(c\) at once; substitute the other two points into \(y = ax^2 + bx + c\) and solve the pair of equations.
- Turning point given: write \(y = (x - h)^2 + k\) (when \(a = 1\)), expand and match with \(y = x^2 + bx + c\).
- How many solutions? For \(ax^2 + bx + c = k\), compare \(k\) with the \(y\)-value of the turning point (Example 4).
- A condition on the turning point (on the \(x\)-axis, on a given line): complete the square with the letter in place, write the turning point in terms of the letter, then apply the condition.
(i) complete a table of values for \(x = -4\) to \(2\) and plot the graph
(ii) state the turning point, the axis of symmetry, the intercepts and the range.
(i) Square first, then add \(2x\) and subtract \(3\). For \(x = -4\): \(16 - 8 - 3 = 5\).
| \(x\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
| \(y\) | \(5\) | \(0\) | \(-3\) | \(-4\) | \(-3\) | \(0\) | \(5\) |
(ii) The \(y\)-values repeat either side of \(x = -1\), where \(y\) is least.
| \(\text{turning point}\) | \(\) | \(\boldsymbol{(-1, -4)}\text{, a minimum}\) |
| \(\text{axis}\) | \(\) | \(\boldsymbol{x = -1}\) |
| \(x\text{-intercepts}\) | \(\) | \(\boldsymbol{x = -3,\ x = 1}\) |
| \(y\text{-intercept}\) | \(\) | \(\boldsymbol{-3}\) |
| \(\text{range}\) | \(\) | \(\boldsymbol{y \ge -4}\) |
\(a = -2\), \(b = 8\), \(c = -5\). Keep the brackets when substituting.
| \(x\) | \(=\) | \(-\dfrac{b}{2a} = -\dfrac{8}{2 \times (-2)} = \mathbf{2}\) |
| \(y\) | \(=\) | \(-2(2)^2 + 8(2) - 5\) |
| \(\) | \(=\) | \(-8 + 16 - 5 = 3\) |
\(a \lt 0\), so the parabola is concave down and \(\mathbf{(2, 3)}\) is a maximum turning point. The range is \(\boldsymbol{y \le 3}\).
The \(y\)-intercept \((0, -5)\) is \(2\) units left of the axis \(x = 2\), so its mirror image is \(2\) units right: \(\mathbf{(4, -5)}\). Check: \(-2(16) + 32 - 5 = -5\).
(i) Explain why its axis of symmetry is \(x = 3\).
(ii) Find \(a\), \(b\) and \(c\), and the turning point.
(i) \((1, 2)\) and \((5, 2)\) have the same \(y\)-value, so they are mirror images: the axis is halfway, \(x = \dfrac{1 + 5}{2} = 3\).
(ii) \((0, 7)\) gives \(c = 7\). Substitute the other two points:
| \(a + b + 7\) | \(=\) | \(2 \quad\Rightarrow\quad a + b = -5\) |
| \(25a + 5b + 7\) | \(=\) | \(2 \quad\Rightarrow\quad 5a + b = -1\) |
Subtract the first equation from the second: \(4a = 4\), so \(a = 1\) and \(b = -6\).
| \(y\) | \(=\) | \(\boldsymbol{x^2 - 6x + 7}\) |
| \(x = 3:\ y\) | \(=\) | \(9 - 18 + 7 = -2\) |
Turning point \(\mathbf{(3, -2)}\). Check with \(-\dfrac{b}{2a} = \dfrac{6}{2} = 3\).
(ii) The parabola \(y = x^2 + 4x\) has turning point \((-2, -4)\). Use horizontal lines \(y = k\) to find the values of \(k\) for which \(x^2 + 4x = k\) has two, one or no solutions.
(i) Complete the square with \(k\) in place. Half of \(k\) is \(\dfrac{k}{2}\).
| \(y\) | \(=\) | \(\left(x + \dfrac{k}{2}\right)^2 + 25 - \dfrac{k^2}{4}\) |
The turning point is \(\left(-\dfrac{k}{2},\ 25 - \dfrac{k^2}{4}\right)\). Touching the \(x\)-axis means the turning point is on it, so its \(y\)-value is \(0\):
| \(25 - \dfrac{k^2}{4}\) | \(=\) | \(0\) |
| \(k^2\) | \(=\) | \(100\) |
| \(k\) | \(=\) | \(\boldsymbol{10} \text{ or } \boldsymbol{-10}\) |
\(k = 10\) touches at \(x = -5\); \(k = -10\) touches at \(x = 5\).
(ii) The solutions of \(x^2 + 4x = k\) are where the line \(y = k\) meets the parabola. The lowest point of the parabola is at \(y = -4\).
| \(\text{two solutions}\) | \(\) | \(\boldsymbol{k \gt -4}\) |
| \(\text{one solution}\) | \(\) | \(\boldsymbol{k = -4}\ (x = -2)\) |
| \(\text{no solutions}\) | \(\) | \(\boldsymbol{k \lt -4}\) |
Common pitfalls
Frequently asked questions
How do you find the turning point of a parabola?
Find the axis of symmetry with \(x = -\dfrac{b}{2a}\) (or halfway between the \(x\)-intercepts), then substitute that \(x\)-value to get the \(y\)-value.
How can you tell if a parabola is concave up or down?
Look at the coefficient of \(x^2\). If it is positive the parabola is concave up with a minimum turning point; if it is negative it is concave down with a maximum turning point.
How do you find the axis of symmetry from a table of values?
Find two \(x\)-values that give the same \(y\)-value. The axis is halfway between them.
What is the range of a parabola?
All the \(y\)-values it reaches. If the turning point is \((h, k)\), the range is \(y \ge k\) for a concave up parabola and \(y \le k\) for a concave down one.
What does the value of a do to y = ax^2?
It sets the width. A larger positive \(a\) makes the parabola narrower, a value between \(0\) and \(1\) makes it wider, and a negative \(a\) turns it upside down.