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Year 10 Maths - 10 Parabolas and Other Graphs

Exploring Parabolas

20 practice questions 1 video lesson Theory + worked examples
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Theory

The graph of a quadratic \(y = ax^2 + bx + c\) is a parabola. This Year 10 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 also shows how \(a\) changes the shape, how to find the axis from \(x = -\dfrac{b}{2a}\) or from symmetry, and how to find 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 \(y = x^2\).

Features of a parabolaA concave up parabola crossing the x-axis at two x-intercepts and the y-axis at the y-intercept. Its lowest point, the vertex, lies on the dashed vertical axis of symmetry. x y O x-intercept x-intercept y-intercept vertex axis of symmetry
The axis of symmetry passes through the turning point and halfway between the \(x\)-intercepts.
The effect of a on y = a x squaredThree parabolas with turning point at the origin. y = 2x squared is narrow, y = x squared is dashed, and y = one quarter x squared is wide. x y O y = 2xΒ² y = xΒ² y = ΒΌxΒ²
In \(y = ax^2\), a larger \(a\) gives a narrower parabola and a smaller positive \(a\) a wider one. All three have turning point \((0, 0)\).

Axis of symmetry of \(y = ax^2 + bx + c\), and the axis from two \(x\)-intercepts \(p\) and \(q\):

\[x = -\dfrac{b}{2a} \qquad x = \dfrac{p + q}{2}\]
x=−b2a

Completing the square when \(a = 1\) gives the turning point directly:

\[x^2 + bx + c = \left(x + \dfrac{b}{2}\right)^2 + c - \dfrac{b^2}{4} \qquad \text{turning point } \left(-\dfrac{b}{2},\ c - \dfrac{b^2}{4}\right)\]
Turning point from the axis. For \(y = x^2 - 8x + 3\), the axis is \(x = -\dfrac{-8}{2 \times 1} = 4\). Substitute: \(y = 16 - 32 + 3 = -13\), so the turning point is \((4, -13)\) and the range is \(y \ge -13\).
Symmetry. Two points with the same \(y\)-value are mirror images in the axis. If \((-2, 9)\) and \((6, 9)\) are on a parabola, its axis is \(x = \dfrac{-2 + 6}{2} = 2\). A point \(3\) units left of the axis has a partner \(3\) units right of it, at the same height.
Number of \(x\)-intercepts. Concave up with the turning point above the \(x\)-axis: none. On the axis: one (the parabola touches it). Below: two. Reverse "above" and "below" for concave down.

How to sketch a parabola

  1. Concavity: read the sign of \(a\).
  2. \(y\)-intercept: \(y = c\).
  3. \(x\)-intercepts: let \(y = 0\) and solve, by factorising if you can.
  4. Axis and turning point: \(x = -\dfrac{b}{2a}\) or the midpoint of the \(x\)-intercepts; substitute to get the \(y\)-value.
  5. 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

  1. The lowest (or highest) point is the turning point; the axis is \(x =\) its \(x\)-coordinate.
  2. In a table, look for the \(y\)-values that repeat: the axis is halfway between their \(x\)-values. The \(x\)-intercepts are where \(y = 0\).
  3. Use the turning point to state the range.

Finding unknown coefficients

  1. 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.
  2. Turning point given: write \(y = (x - h)^2 + k\) (when \(a = 1\)), expand and match with \(y = x^2 + bx + c\).
  3. 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.
Example 1 β€” From a table of values
For \(y = x^2 + 2x - 3\):
(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.
Solution

(i) Square first, then add \(2x\) and subtract \(3\). For \(x = -4\): \(16 - 8 - 3 = 5\).

\(x\)\(-4\)\(-3\)\(-2\)\(-1\)\(0\)\(1\)\(2\)
\(y\)\(5\)\(0\)\(-3\)\(-4\)\(-3\)\(0\)\(5\)
Graph of y = x squared + 2x minus 3 from a table of valuesSeven plotted points from x = -4 to 2 joined by a smooth concave up parabola with turning point (-1, -4), x-intercepts -3 and 1, y-intercept -3 and dashed axis of symmetry x = -1. x y O βˆ’4 βˆ’2 2 βˆ’4 βˆ’2 2 4 x = βˆ’1 (βˆ’1, βˆ’4)

(ii) The \(y\)-values repeat either side of \(x = -1\), where \(y\) is least.

\(\text{turning point}\)\(\)\(\boldsymbol{(-1, -4)}\text{, a minimum}\)
\(\text{axis}\)\(\)\(\boldsymbol{x = -1}\)
\(x\text{-intercepts}\)\(\)\(\boldsymbol{x = -3,\ x = 1}\)
\(y\text{-intercept}\)\(\)\(\boldsymbol{-3}\)
\(\text{range}\)\(\)\(\boldsymbol{y \ge -4}\)
Example 2 β€” A concave down parabola
For \(y = -2x^2 + 8x - 5\), find the axis of symmetry, the turning point and the range. Then use symmetry to find another point with the same \(y\)-value as the \(y\)-intercept.
Solution

\(a = -2\), \(b = 8\), \(c = -5\). 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: \(-2(16) + 32 - 5 = -5\).

(2,3)
Example 3 β€” Equation from three points
A parabola \(y = ax^2 + bx + c\) passes through \((0, 7)\), \((1, 2)\) and \((5, 2)\).
(i) Explain why its axis of symmetry is \(x = 3\).
(ii) Find \(a\), \(b\) and \(c\), and the turning point.
Solution

(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 \(-\dfrac{b}{2a} = \dfrac{6}{2} = 3\).

Parabola through (0, 7), (1, 2) and (5, 2)Concave up parabola y = x squared - 6x + 7 with dashed axis x = 3 halfway between the points (1, 2) and (5, 2), y-intercept 7 and vertex (3, -2). x y O 1 3 5 x = 3 (0, 7) (1, 2) (5, 2) (3, βˆ’2)
Example 4 β€” Unknown coefficients
(i) The parabola \(y = x^2 + kx + 25\) touches the \(x\)-axis. Find the possible values of \(k\).
(ii) The parabola \(y = x^2 + bx + c\) has turning point \((-2, 5)\). Find \(b\) and \(c\).
Solution

(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) With \(a = 1\) and turning point \((-2, 5)\), the equation is \(y = (x + 2)^2 + 5\). Expand and match:

\(y\)\(=\)\(x^2 + 4x + 4 + 5 = x^2 + 4x + 9\)

So \(\boldsymbol{b = 4}\) and \(\boldsymbol{c = 9}\).

Common pitfalls

Watch the sign in \(-\dfrac{b}{2a}\). For \(y = -x^2 + 6x\), \(x = -\dfrac{6}{2(-1)} = 3\), not \(-3\). Substitute with brackets.
Square before you change the sign. For \(y = 7 - x^2\) at \(x = -3\): \(7 - (-3)^2 = 7 - 9 = -2\), not \(7 + 9\).
The axis is a line. Write the axis of symmetry as an equation, such as \(x = 3\), not just \(3\) or the point \((3, 0)\).
Maximum or minimum? Concave up gives a minimum and range \(y \ge k\); concave down gives a maximum and range \(y \le k\).

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.