Parabolas Using The Quadratic Formula
Every question with a fully worked solution.
- Parabolas Using The Quadratic Formula - Video - Sketching Parabolas by using the Quadratic Formula Watch
Theory
When \(y = ax^2 + bx + c\) does not factorise, the quadratic formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) gives its \(x\)-intercepts, and the discriminant \(\Delta = b^2 - 4ac\) tells you how many there are. This Year 10A maths page shows how to find exact and decimal intercepts, sketch the parabola with its turning point, and use the discriminant to find unknown coefficients.
The \(x\)-intercepts of the parabola \(y = ax^2 + bx + c\) are the solutions of \(ax^2 + bx + c = 0\). When the quadratic does not factorise, the quadratic formula finds them. The coefficients \(a\), \(b\) and \(c\) include their signs: in \(y = 2x^2 - 7x - 3\), \(a = 2\), \(b = -7\) and \(c = -3\).
The \(\pm\) sign gives both \(x\)-intercepts at once. They are the same distance either side of the axis of symmetry \(x = -\dfrac{b}{2a}\), which is the first part of the formula.
The expression under the square root, \(\Delta = b^2 - 4ac\), is the discriminant. If \(\Delta \gt 0\) the parabola has two \(x\)-intercepts, if \(\Delta = 0\) it touches the \(x\)-axis once (at the turning point), and if \(\Delta \lt 0\) it does not meet the \(x\)-axis.
An exact intercept keeps the surd, such as \(\dfrac{-5 \pm \sqrt{17}}{2}\). A decimal approximation rounds each one separately, such as \(-0.44\) and \(-4.56\). These are what you plot.
The \(x\)-intercepts of \(y = ax^2 + bx + c\), and the discriminant:
| Discriminant | Number of real solutions | Graph of \(y=ax^2+bx+c\) |
|---|---|---|
| \(\Delta\gt 0\) | two (different) | crosses the \(x\)-axis twice |
| \(\Delta=0\) | one (repeated) | touches the \(x\)-axis |
| \(\Delta\lt 0\) | none | does not meet the \(x\)-axis |
How to sketch a parabola with the quadratic formula
- Set \(y = 0\): the \(x\)-intercepts solve \(ax^2 + bx + c = 0\). Clear any decimals or fractions.
- Write down \(a\), \(b\) and \(c\) with their signs.
- Work out \(\Delta=b^2-4ac\) on its own line. Use brackets for a negative \(b\): \((-6)^2=36\). If \(\Delta\lt 0\), stop: there are no real solutions.
- Substitute into \(x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\).
- Simplify the surd and cancel any common factor for an exact answer, or use a calculator for each sign separately for a decimal answer.
- Turning point: \(x = -\dfrac{b}{2a}\); substitute for \(y\). \(y\)-intercept: \(c\).
- Sketch through the decimal values of the intercepts, the turning point and the \(y\)-intercept. In a context, reject impossible values such as a negative distance.
Questions about the discriminant
- How many \(x\)-intercepts? Work out \(\Delta\) and use its sign.
- Touches the \(x\)-axis: set \(\Delta = 0\) and solve for the unknown. The point of contact is the turning point, \(x = -\dfrac{b}{2a}\).
- Two intercepts: solve \(\Delta \gt 0\). For \(k^2 - 36 \gt 0\), the solutions lie outside \(\pm 6\): \(k \lt -6\) or \(k \gt 6\).
- Always two intercepts: write \(\Delta\) as a square plus a positive number, which can never be \(0\) or negative.
- Above or below the axis: if \(\Delta \lt 0\), the whole parabola is above the \(x\)-axis when \(a \gt 0\) and below it when \(a \lt 0\).
(i) \(y = x^2 - 3x - 7\)
(ii) \(y = 9x^2 + 6x + 1\)
(iii) \(y = 2x^2 - x + 3\)
| \(\text{(i) }\Delta\) | \(=\) | \((-3)^2 - 4(1)(-7) = 9 + 28 = 37\) |
| \(\text{(ii) }\Delta\) | \(=\) | \(6^2 - 4(9)(1) = 36 - 36 = 0\) |
| \(\text{(iii) }\Delta\) | \(=\) | \((-1)^2 - 4(2)(3) = 1 - 24 = -23\) |
(i) \(\Delta \gt 0\): two \(x\)-intercepts.
(ii) \(\Delta = 0\): one. The parabola touches the \(x\)-axis at its turning point, \(x = -\dfrac{6}{18} = -\dfrac{1}{3}\).
(iii) \(\Delta \lt 0\): none. Since \(a = 2 \gt 0\), the parabola is concave up and lies entirely above the \(x\)-axis.
(i) find the \(x\)-intercepts exactly and correct to \(2\) decimal places
(ii) find the turning point and the \(y\)-intercept, and sketch the parabola.
(i) \(a = 1\), \(b = 5\), \(c = 2\). \(\Delta = 25 - 8 = 17\).
| \(x\) | \(=\) | \(\dfrac{-5 \pm \sqrt{17}}{2}\) |
| \(\) | \(\approx\) | \(\boldsymbol{-0.44} \text{ or } \boldsymbol{-4.56}\) |
(ii) The axis is \(x = -\dfrac{5}{2} = -2.5\), and \(y = 6.25 - 12.5 + 2 = -4.25\). Turning point \(\mathbf{(-2.5, -4.25)}\), a minimum. \(y\)-intercept \(\mathbf{2}\).
\(a = -2\), \(b = 7\), \(c = -2\). \(\Delta = 49 - 16 = 33\).
| \(x\) | \(=\) | \(\dfrac{-7 \pm \sqrt{33}}{-4}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{7 \mp \sqrt{33}}{4}}\) |
| \(\) | \(\approx\) | \(\boldsymbol{0.31} \text{ or } \boldsymbol{3.19}\) |
Axis \(x = -\dfrac{7}{2(-2)} = 1.75\); \(y = -6.125 + 12.25 - 2 = 4.125\). Turning point \(\mathbf{(1.75, 4.125)}\), a maximum. \(y\)-intercept \(\mathbf{-2}\).
Dividing by a negative \(2a\) swaps \(\pm\) to \(\mp\); the two intercepts are the same either way.
(ii) Show that \(y = x^2 + kx + k - 2\) has two \(x\)-intercepts for every value of \(k\).
(i) \(a = 1\), \(b = k\), \(c = 9\). Two intercepts need \(\Delta \gt 0\).
| \(\Delta\) | \(=\) | \(k^2 - 36\) |
| \(k^2 - 36\) | \(\gt\) | \(0\) |
| \(k^2\) | \(\gt\) | \(36\) |
\(k^2 \gt 36\) when \(k\) is more than \(6\) away from \(0\): \(\boldsymbol{k \lt -6}\) or \(\boldsymbol{k \gt 6}\). (At \(k = \pm 6\) the parabola touches the axis.)
(ii) \(a = 1\), \(b = k\), \(c = k - 2\). Complete the square on \(\Delta\):
| \(\Delta\) | \(=\) | \(k^2 - 4(k - 2)\) |
| \(\) | \(=\) | \(k^2 - 4k + 8\) |
| \(\) | \(=\) | \((k - 2)^2 + 4\) |
\((k - 2)^2 \ge 0\), so \(\Delta \ge 4 \gt 0\) for every \(k\): there are always two \(x\)-intercepts.
Common pitfalls
Frequently asked questions
What is the quadratic formula?
For \(ax^2+bx+c=0\), \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\). It solves every quadratic equation, including ones that do not factorise.
What does the discriminant tell you?
The discriminant \(b^2-4ac\) tells you the number of real solutions: two if it is positive, one if it is zero, and none if it is negative.
When should I use the formula instead of factorising?
Factorise when the numbers work out easily. Use the formula when the quadratic does not factorise, when the question asks for surds or decimals, or when you cannot see the factors quickly.
How do I simplify an answer like (6 plus or minus 2 root 5) over 2?
Every term in the numerator has a factor of \(2\), so divide each one by \(2\) to get \(3\pm\sqrt{5}\). Simplify the surd before cancelling.
Why is one solution rejected in word problems?
The formula gives every solution of the equation, but the situation may not allow some of them. A length or a time after the start cannot be negative, and a box cannot have a side of negative length.