Revision (EMBGK)
Functions of the form \(y = a x^2 + q\)
Functions of the general form \(y=a{x}^{2}+q\) are called parabolic functions, where \(a\) and \(q\) are constants.
The effects of \(a\) and \(q\) on \(f(x) = ax^2 + q\):
-
The effect of \(q\) on vertical shift
-
For \(q>0\), \(f(x)\) is shifted vertically upwards by \(q\) units.
The turning point of \(f(x)\) is above the \(x\)-axis.
-
For \(q<0\), \(f(x)\) is shifted vertically downwards by \(q\) units.
The turning point of \(f(x)\) is below the \(x\)-axis.
-
\(q\) is also the \(y\)-intercept of the parabola.
-
-
The effect of \(a\) on shape
-
For \(a>0\); the graph of \(f(x)\) is a “smile” and has a minimum turning point \((0;q)\). As the value of \(a\) becomes larger, the graph becomes narrower.
As \(a\) gets closer to \(\text{0}\), \(f(x)\) becomes wider.
-
For \(a<0\); the graph of \(f(x)\) is a “frown” and has a maximum turning point \((0;q)\). As the value of \(a\) becomes smaller, the graph becomes narrower.
As \(a\) gets closer to \(\text{0}\), \(f(x)\) becomes wider.
-
| \(a<0\) | \(a>0\) | |
| \(q>0\) |  |  |
| \(q=0\) |  |  |
| \(q<0\) |  |  |
Revision
Textbook Exercise 5.1
On separate axes, accurately draw each of the following functions.
- Use tables of values if necessary.
- Use graph paper if available.
\(y_1 = x^2\)
\(y_2 = \frac{1}{2}x^2\)
\(y_3 = -x^2 - 1\)
\(y_4 = -2x^2 + 4\)
Use your sketches of the functions given above to complete the following table (the first column has been completed as an example):
| \(y_1\) | \(y_2\) | \(y_3\) | \(y_4\) | |
| value of \(q\) | \(q = 0\) | |||
| effect of \(q\) | \(y_{\text{int}} = 0\) | |||
| value of \(a\) | \(a = 1\) | |||
| effect of \(a\) | standard parabola | |||
| turning point | \((0;0)\) | |||
| axis of symmetry | \(x = 0\) (\(y\)-axis) | |||
| domain | \(\{x: x \in \mathbb{R} \}\) | |||
| range | \(\{y: y \geq 0\}\) |
| \(y_1\) | \(y_2\) | \(y_3\) | \(y_4\) | |
| value of \(q\) | \(q = \text{0}\) | \(q = \text{0}\) | \(q = -\text{1}\) | \(q = \text{4}\) |
| effect of \(q\) | \(y_{\text{int}} = 0\) | \(y_{\text{int}} = 0\) | \(y_{\text{int}} = -1\), shifts \(\text{1}\) unit down | \(y_{\text{int}} = 4\), shifts \(\text{4}\) unit up |
| value of \(a\) | \(a = 1\) | \(a = \frac{1}{2}\) | \(a = -1\) | \(a = -2\) |
| effect of \(a\) | standard parabola | smile, min. TP | frown, max. TP | frown, max. TP |
| turning point | \((0;0)\) | \((0;0)\) | \((0;-1)\) | \((0;4)\) |
| axis of symmetry | \(x = 0\) (\(y\)-axis) | \(x = 0\) (\(y\)-axis) | \(x = 0\) (\(y\)-axis) | \(x = 0\) (\(y\)-axis) |
| domain | \(\{x: x \in \mathbb{R} \}\) | \(\{x: x \in \mathbb{R} \}\) | \(\{x: x \in \mathbb{R} \}\) | \(\{x: x \in \mathbb{R} \}\) |
| range | \(\{y: y \geq 0\}\) | \(\{y: y \geq 0\}\) | \(\{y: y \leq -1\}\) | \(\{y: y \leq 4\}\) |
Functions of the form \(y=a{\left(x+p\right)}^{2}+q\) (EMBGM)
We now consider parabolic functions of the form \(y=a{\left(x+p\right)}^{2}+q\) and the effects of parameter \(p\).
The effects of \(a\), \(p\) and \(q\) on a parabolic graph
-
On the same system of axes, plot the following graphs:
- \(y_1 = x^2\)
- \(y_2 = (x - 2)^2\)
- \(y_3 = (x - 1)^2\)
- \(y_4 = (x + 1)^2\)
- \(y_5 = (x + 2)^2\)
Use your sketches of the functions above to complete the following table:
\(y_1\) \(y_2\) \(y_3\) \(y_4\) \(y_5\) \(x\)-intercept(s) \(y\)-intercept turning point axis of symmetry domain range effect of \(p\) -
On the same system of axes, plot the following graphs:
- \(y_1 = x^2 + 2\)
- \(y_2 = (x - 2)^2 - 1\)
- \(y_3 = (x - 1)^2 + 1\)
- \(y_4 = (x + 1)^2 + 1\)
- \(y_5 = (x + 2)^2 - 1\)
Use your sketches of the functions above to complete the following table:
\(y_1\) \(y_2\) \(y_3\) \(y_4\) \(y_5\) \(x\)-intercept(s) \(y\)-intercept turning point axis of symmetry domain range effect of \(q\) -
Consider the three functions given below and answer the questions that follow:
- \(y_1 = (x - 2)^2 + 1\)
- \(y_2 = 2(x - 2)^2 + 1\)
- \(y_3 = -\frac{1}{2}(x - 2)^2 + 1\)
-
What is the value of \(a\) for \(y_2\)?
-
Does \(y_1\) have a minimum or maximum turning point?
-
What are the coordinates of the turning point of \(y_2\)?
-
Compare the graphs of \(y_1\) and \(y_2\). Discuss the similarities and differences.
-
What is the value of \(a\) for \(y_3\)?
-
Will the graph of \(y_3\) be narrower or wider than the graph of \(y_1\)?
-
Determine the coordinates of the turning point of \(y_3\).
-
Compare the graphs of \(y_1\) and \(y_3\). Describe any differences.
The effect of the parameters on \(y = a(x + p)^2 + q\)
The effect of \(p\) is a horizontal shift because all points are moved the same distance in the same direction (the entire graph slides to the left or to the right).
-
For \(p>0\), the graph is shifted to the left by \(p\) units.
-
For \(p<0\), the graph is shifted to the right by \(p\) units.
The value of \(p\) also affects whether the turning point is to the left of the \(y\)-axis \(\left(p>0\right)\) or to the right of the \(y\)-axis \(\left(p<0\right)\). The axis of symmetry is the line \(x = -p\).
The effect of \(q\) is a vertical shift. The value of \(q\) affects whether the turning point of the graph is above the \(x\)-axis \(\left(q>0\right)\) or below the \(x\)-axis \(\left(q<0\right)\).
The value of \(a\) affects the shape of the graph. If \(a<0\), the graph is a “frown” and has a maximum turning point. If \(a>0\) then the graph is a “smile” and has a minimum turning point. When \(a = 0\), the graph is a horizontal line \(y = q\).
| \(p>0\) | \(p<0\) | |||
| \(a<0\) | \(a>0\) | \(a<0\) | \(a>0\) | |
| \(q>0\) |  |  |  |  |
| \(q<0\) |  |  |  |  |
Discovering the characteristics
For functions of the general form \(f(x) = y = a(x + p)^2 + q\):
Domain and range
The domain is \(\left\{x:x\in ℝ\right\}\) because there is no value of \(x\) for which \(f(x)\) is undefined.
The range of \(f(x)\) depends on whether the value for \(a\) is positive or negative. If \(a>0\) we have: \[\begin{array}{r@{\;}c@{\;}l@{\quad}l} (x + p)^2 & \geq & 0 & (\text{perfect square is always positive}) \\ \therefore a(x + p)^2 & \geq & 0 & (a \text{ is positive}) \\ \therefore a(x + p)^2 + q & \geq & q & \\ \therefore f(x) & \geq & q & \end{array}\]
The range is therefore \(\{ y: y \geq q, y \in \mathbb{R} \}\) if \(a > 0\). Similarly, if \(a < 0\), the range is \(\{ y: y \leq q, y \in \mathbb{R} \}\).
Worked example 1: Domain and range
State the domain and range for \(g(x) = -2(x - 1)^2 + 3\).
Determine the domain
The domain is \(\{x: x \in \mathbb{R} \}\) because there is no value of \(x\) for which \(g(x)\) is undefined.
Determine the range
The range of \(g(x)\) can be calculated from: \begin{align*} (x - 1)^2& \geq 0 \\ -2(x - 1)^2& \leq 0 \\ -2(x - 1)^2 + 3 & \leq 3 \\ g(x) & \leq 3 \end{align*} Therefore the range is \(\{g(x): g(x) \leq 3 \}\) or in interval notation \((-\infty; 3]\).
Notice in the example above that it helps to have the function in the form \(y = a(x + p)^2 + q\).
We use the method of completing the square to write a quadratic function of the general form \(y = ax^2 + bx +c\) in the form \(y = a(x + p)^2 + q\) (see Chapter \(\text{2}\)).
Domain and range
Textbook Exercise 5.2
Give the domain and range for each of the following functions:
\(f(x) = (x-4)^2 - 1\)
\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y \geq -1, y\in \mathbb{R} \right \} \end{align*}
\(g(x) = -(x-5)^2 + 4\)
\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y \leq 4, y\in \mathbb{R} \right \} \end{align*}
\(h(x) = x^2 -6x +9\)
\begin{align*} h(x) &= x^2 -6x +9 \\ &= (x - 3)^2 \\ \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y \geq 0, y\in \mathbb{R} \right \} \end{align*}
\(j(x) = -2(x+1)^2\)
\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y \leq 0, y\in \mathbb{R} \right \} \end{align*}
\(k(x) = -x^2 + 2x - 3\)
\begin{align*} k(x) &= -x^2 + 2x - 3 \\ &= -(x - 1)^2 - 2 \\ \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y \leq 2, y\in \mathbb{R} \right \} \end{align*}
Intercepts
The \(y\)-intercept:
Every point on the \(y\)-axis has an \(x\)-coordinate of \(\text{0}\), therefore to calculate the \(y\)-intercept we let \(x=0\).
For example, the \(y\)-intercept of \(g(x) = (x - 1)^2 + 5\) is determined by setting \(x=0\): \begin{align*} g(x) &= (x - 1)^2 + 5 \\ g(0) &= (0 - 1)^2 + 5 \\ &= 6 \end{align*} This gives the point \((0;6)\).
The \(x\)-intercept:
Every point on the \(x\)-axis has a \(y\)-coordinate of \(\text{0}\), therefore to calculate the \(x\)-intercept we let \(y=0\).
For example, the \(x\)-intercept of \(g(x) = (x - 1)^2 + 5\) is determined by setting \(y=0\): \begin{align*} g(x) &= (x - 1)^2 + 5 \\ 0 &= (x - 1)^2 + 5 \\ -5 &= (x - 1)^2 \end{align*} which has no real solutions. Therefore, the graph of \(g(x)\) lies above the \(x\)-axis and does not have any \(x\)-intercepts.
Intercepts
Textbook Exercise 5.3
Determine the \(x\)- and \(y\)-intercepts for each of the following functions:
\(f(x) = (x+4)^2 - 1\)
\begin{align*} \text{For } x=0 \quad y &= (0+4)^2 - 1 \\ &= 16 - 1 \\ \therefore & (0;15) \\ \text{For } y=0 \quad 0 &= (x+4)^2 - 1 \\ &= x^2 + 8x + 16 - 1 \\ &= x^2 + 8x + 15 \\ &= (x + 5)(x + 3) \\ x=-5 &\text{ or } x=-3 \\ \therefore (-5;0) &\text{ and } (-3;0) \end{align*}
\(g(x) = \text{16} - 8x + x^2\)
\begin{align*} \text{For } x=0 \quad y &=16 \\ \therefore & (0;16) \\ \text{For } y=0 \quad 0 &= 16 - 8x + x^2 \\ &= x^2 - 8x + 16 \\ &= (x -4)^2 \\ \therefore x &=4 \\ \therefore & (4;0) \end{align*}
\(h(x) = -x^2 +4x-3\)
\begin{align*} \text{For } x=0 \quad y &=-3 \\ \therefore & (0;-3) \\ \text{For } y=0 \quad 0 &= -x^2 +4x-3 \\ &= - (x^2 -4x+3) \\ &= -(x - 3)(x - 1) \\ x=3 &\text{ or } x=1 \\ \therefore (1;0) &\text{ and } (3;0) \end{align*}
\(j(x) = 4(x-3)^2 -1\)
\begin{align*} \text{For } x=0 \quad y &= 4(0-3)^2 -1 \\ &= 36 - 1 \\ \therefore & (0;35) \\ \text{For } y=0 \quad 0 &= 4(x-3)^2 -1 \\ &= 4(x^2 - 6x + 9) -1 \\ &= 4x^2 -24x + 36 - 1 \\ &= 4x^2 -36x + 35 \\ &= (2x + 5)(2x + 7) \\ x=-\frac{5}{2} &\text{ or } x=-\frac{7}{2} \\ \therefore (-\frac{5}{2};0) &\text{ and } (-\frac{7}{2};0) \end{align*}
\(k(x) = 4(x-3)^2 +1\)
\begin{align*} \text{For } x=0 \quad y &= 4(0-3)^2 +1 \\ &= 36 +1 \\ \therefore & (0;37) \\ \text{For } y=0 \quad 0 &= 4(x-3)^2 +1 \\ &= 4(x^2 - 6x + 9) +1 \\ &= 4x^2 -24x + 36 + 1 \\ &= 4x^2 -36x + 37 \\ \therefore x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ & = \frac{24 \pm \sqrt{(-24)^2 - 4(4)(37)}}{2(4)} \\ & = \frac{576 \pm \sqrt{576 - 592}}{8} \\ & = \frac{576 \pm \sqrt{-16}}{8} \\ \therefore &\text{ no real solution } \end{align*}
\(l(x) = 2x^2 - 3x -4\)
\begin{align*} \text{For } x=0 \quad y &=-4 \\ \therefore & (0;-4) \\ \text{For } y=0 \quad 0 &= 2x^2 - 3x -4 \\ &= 2x^2 - 3x -4 \\ \therefore x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ & = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-4)}}{2(2)} \\ & = \frac{3 \pm \sqrt{ 9 + 32}}{4} \\ & = \frac{3 \pm \sqrt{41}}{4} \\ x=\frac{3 - \sqrt{41}}{4} &\text{ or } x=\frac{3 + \sqrt{41}}{4} \\ \therefore (-\text{0,85};0) &\text{ and } (\text{2,35};0) \end{align*}
Turning point
The turning point of the function \(f(x) = a(x+p)^2 + q\) is determined by examining the range of the function:
-
If \(a > 0\), \(f(x)\) has a minimum turning point and the range is \([q;\infty)\):
The minimum value of \(f(x)\) is \(q\).
If \(f(x) = q\), then \(a(x+p)^2 = 0\), and therefore \(x = -p\).
This gives the turning point \((-p;q)\).
-
If \(a < 0\), \(f(x)\) has a maximum turning point and the range is \((-\infty;q]\):
The maximum value of \(f(x)\) is \(q\).
If \(f(x) = q\), then \(a(x+p)^2 = 0\), and therefore \(x = -p\).
This gives the turning point \((-p;q)\).
Therefore the turning point of the quadratic function \(f(x) = a(x+p)^2 + q\) is \((-p;q)\).
Alternative form for quadratic equations:
We can also write the quadratic equation in the form
\[y = a(x - p)^2 +q\]The effect of \(p\) is still a horizontal shift, however notice that:
-
For \(p>0\), the graph is shifted to the right by \(p\) units.
-
For \(p<0\), the graph is shifted to the left by \(p\) units.
The turning point is \((p;q)\) and the axis of symmetry is the line \(x = p\).
Worked example 2: Turning point
Determine the turning point of \(g(x) = 3x^2 - 6x - 1\).
Write the equation in the form \(y = a(x+p)^2 + q\)
We use the method of completing the square: \begin{align*} g(x )&= 3x^2 - 6x - 1 \\ &= 3(x^2 - 2x) - 1 \\ &= 3 \left( (x-1)^2 - 1 \right) -1 \\ &= 3(x-1)^2 - 3 -1 \\ &= 3(x-1)^2 - 4 \end{align*}
Determine turning point \((-p;q)\)
From the equation \(g(x) = 3(x-1)^2 - 4\) we know that the turning point for \(g(x)\) is \((1;-4)\).
Worked example 3: Turning point
- Show that the \(x\)-value for the turning point of \(h(x) = ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\).
- Hence, determine the turning point of \(k(x) = 2 - 10x + 5x^2\).
Write the equation in the form \(y = a(x+p)^2 + q\) and show that \(p = \frac{b}{2a}\)
We use the method of completing the square: \begin{align*} h(x)&= ax^2 + bx + c \\ &= a \left( x^2 + \frac{b}{a}x + \frac{c}{a} \right) \end{align*}
Take half the coefficient of the \(x\) term and square it; then add and subtract it from the expression. \begin{align*} h(x) &= a \left( x^2 + \frac{b}{a}x + \left( \frac{b}{2a} \right)^2 - \left( \frac{b}{2a} \right)^2 + \frac{c}{a} \right) \\ &= a \left( \left(x + \frac{b}{2a} \right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} \right) \\ &= a \left( \left(x + \frac{b}{2a} \right)^2 - \frac{b^2 -4ac}{4a^2} \right) \\ &= a \left(x + \frac{b}{2a} \right)^2 - \frac{b^2 -4ac}{4a} \end{align*} From the above we have that the turning point is at \(x = -p = - \frac{b}{2a}\) and \(y = q = - \frac{b^2 -4ac}{4a}\).
Determine the turning point of \(k(x)\)
Write the equation in the general form \(y = ax^2 + bx + c\). \[k(x) = 5x^2 -10x + 2\] Therefore \(a = 5\); \(b = -10\); \(c = 2\).
Use the results obtained above to determine \(x = - \frac{b}{2a}\): \begin{align*} x &= -\left(\frac{-10}{2(5)}\right) \\ &= 1 \end{align*}
Substitute \(x = 1\) to obtain the corresponding \(y\)-value: \begin{align*} y &= 5x^2 -10x + 2 \\ &= 5(1)^2 -10(1) + 2\\ &= 5 - \text{10} + 2\\ &= -3 \end{align*} The turning point of \(k(x)\) is \((1;-3)\).
Turning points
Textbook Exercise 5.4
Determine the turning point of each of the following:
\(y = x^2 - 6x + 8\)
\begin{align*} y &= x^2 - 6x + 8 \\ &= (x-3)^{2} - \left( \frac{6}{2} \right)^2 + 8 \\ &= (x-3)^{2}-1 \\ \therefore \text{turning point }&= (3;-1) \end{align*}
\(y = -x^2 + 4x - 3\)
\begin{align*} y &= -x^2 + 4x - 3 \\ &= -(x^2 - 4x + 3 \\ &= - \left( (x-2)^{2} - \left( \frac{4}{2} \right)^2 + 3 \right) \\ &= -(x-2)^{2}+1 \\ \therefore \text{turning point } &= (2;1) \end{align*}
\(y = \frac{1}{2}(x + 2)^2 - 1\)
\begin{align*} y &= \frac{1}{2}(x + 2)^2 - 1 \\ \therefore \text{turning point }&= (-2;-1) \end{align*}
\(y = 2x^2 + 2x + 1\)
\begin{align*} y &= 2x^2 + 2x + 1 \\ &= 2 \left( x^2 + x + \frac{1}{2} \right) \\ &= 2 \left( x+ \frac{1}{2} \right )^2+\frac{1}{2} \\ \therefore \text{turning point }&= \left(-\frac{1}{2};\frac{1}{2}\right) \end{align*}
\(y = \text{18} + 6x - 3x^2\)
\begin{align*} y &= -3x^2 + 6x + 18 \\ &= -3(x^2 - 2x - 6) \\ &= -3 \left((x - 1)^2 - 7 \right) \\ &= -3(x-1)^2+21 \\ \therefore \text{turning point }&= (1;21) \end{align*}
\(y = -2[(x + 1)^2 + 3]\)
\begin{align*} y &= -2[(x + 1)^2 + 3] \\ y &= -2(x + 1)^2 - 6 \\ \therefore \text{turning point }&= (-1;-6) \end{align*}
Axis of symmetry
The axis of symmetry for \(f(x)=a{\left(x+p\right)}^{2}+q\) is the vertical line \(x=-p\). The axis of symmetry passes through the turning point \((-p;q)\) and is parallel to the \(y\)-axis.
Axis of symmetry
Textbook Exercise 5.5
Determine the axis of symmetry of each of the following:
\(y = 2x^2 - 5x - \text{18}\)
\begin{align*} y &= 2x^2 - 5x - 18 \\ &= 2(x^2 - \frac{5}{2}x - 9) \\ &= 2\left( \left( x - \frac{5}{2} \right)^2 - \frac{25}{16} - 9 \right) \\ &= 2\left( \left( x - \frac{5}{2} \right)^2 - \frac{169}{16}\right) \\ &= 2\left( x - \frac{5}{4} \right)^2 - \frac{169}{8} \\ \text{Axis of symmetry: }x & =\frac{5}{4} \end{align*}
\(y = 3(x - 2)^2 + 1\)
\begin{align*} y &= 3(x - 2)^2 + 1 \\ \text{Axis of symmetry: } x & = 2 \end{align*}
\(y = 4x - x^2\)
\begin{align*} y &= 4x - x^2 \\ &= -(x^2 - 4x) \\ &= -(x - 2)^2 + 4 \\ \text{Axis of symmetry: } x & = 2 \end{align*}
Write down the equation of a parabola where the \(y\)-axis is the axis of symmetry.
\(y = ax^2 + q\)
Sketching graphs of the form \(f(x)=a{\left(x+p\right)}^{2}+q\)
In order to sketch graphs of the form \(f(x)=a{\left(x+p\right)}^{2}+q\), we need to determine five characteristics:
-
sign of \(a\)
-
turning point
-
\(y\)-intercept
-
\(x\)-intercept(s) (if they exist)
-
domain and range
Worked example 4: Sketching a parabola
Sketch the graph of \(y = -\frac{1}{2}(x + 1)^2 - 3\).
Mark the intercepts, turning point and the axis of symmetry. State the domain and range of the function.
Examine the equation of the form \(y = a(x + p)^2 + q\)
We notice that \(a < 0\), therefore the graph is a “frown” and has a maximum turning point.
Determine the turning point \((-p;q)\)
From the equation we know that the turning point is \((-1; -3)\).
Determine the axis of symmetry \(x = -p\)
From the equation we know that the axis of symmetry is \(x = -1\).
Determine the \(y\)-intercept
The \(y\)-intercept is obtained by letting \(x = 0\): \begin{align*} y &= -\frac{1}{2} \left((0) + 1 \right)^2 - 3\\ &= -\frac{1}{2} - 3\\ &= -3\frac{1}{2} \end{align*} This gives the point \((0;-3\frac{1}{2})\).
Determine the \(x\)-intercepts
The \(x\)-intercepts are obtained by letting \(y = 0\): \begin{align*} 0 &= -\frac{1}{2} \left(x + 1 \right)^2 - 3\\ 3 &= -\frac{1}{2} \left(x + 1 \right)^2 \\ -6 &= \left(x + 1 \right)^2 \end{align*} which has no real solutions. Therefore, there are no \(x\)-intercepts and the graph lies below the \(x\)-axis.
Plot the points and sketch the graph
State the domain and range
Domain: \(\{ x: x \in \mathbb{R} \}\)
Range: \(\{ y: y \leq -3, y \in \mathbb{R} \}\)
Worked example 5: Sketching a parabola
Sketch the graph of \(y = \frac{1}{2}x^2 - 4x + \frac{7}{2}\).
Determine the intercepts, turning point and the axis of symmetry. Give the domain and range of the function.
Examine the equation of the form \(y = ax^2 + bx +c\)
We notice that \(a > 0\), therefore the graph is a “smile” and has a minimum turning point.
Determine the turning point and the axis of symmetry
Check that the equation is in standard form and identify the coefficients.
\[a = \frac{1}{2}; \qquad b = -4; \qquad c = \frac{7}{2}\]
Calculate the \(x\)-value of the turning point using \begin{align*} x &= -\frac{b}{2a} \\ &= -\left(\frac{-4}{2\left( \frac{1}{2} \right)}\right) \\ &= 4 \end{align*} Therefore the axis of symmetry is \(x = 4\).
Substitute \(x = 4\) into the original equation to obtain the corresponding \(y\)-value. \begin{align*} y &= \frac{1}{2}x^2 - 4x + \frac{7}{2} \\ &= \frac{1}{2}(4)^2 - 4(4) + \frac{7}{2} \\ &= 8 -16 +\frac{7}{2} \\ &= -4\frac{1}{2} \end{align*} This gives the point \(\left( 4; -4\frac{1}{2} \right)\).
Determine the \(y\)-intercept
The \(y\)-intercept is obtained by letting \(x = 0\): \begin{align*} y &= \frac{1}{2}(0)^2 - 4(0) + \frac{7}{2}\\ &= \frac{7}{2} \end{align*} This gives the point \(\left(0;\frac{7}{2}\right)\).
Determine the \(x\)-intercepts
The \(x\)-intercepts are obtained by letting \(y = 0\): \begin{align*} 0 &= \frac{1}{2}x^2 - 4x + \frac{7}{2} \\ &= x^2 - 8x + 7 \\ &= (x - 1)(x - 7) \end{align*} Therefore \(x = 1\) or \(x = 7\). This gives the points \((1;0)\) and \((7;0)\).
Plot the points and sketch the graph
State the domain and range
Domain: \(\{ x: x \in \mathbb{R} \}\)
Range: \(\{ y: y \geq -4\frac{1}{2}, y \in \mathbb{R} \}\)
Shifting the equation of a parabola
Carl and Eric are doing their Mathematics homework and decide to check each others answers.
Homework question:
If the parabola \(y = 3x^2 + 1\) is shifted \(\text{2}\) units to the right, determine the equation of the new parabola.
-
Carl's answer:
A shift to the right means moving in the positive \(x\) direction, therefore \(x\) is replaced with \(x + 2\) and the new equation is \(y = 3(x + 2)^2 + 1\).
-
Eric's answer:
We replace \(x\) with \(x - 2\), therefore the new equation is \(y = 3(x - 2)^2 + 1\).
Work together in pairs. Discuss the two different answers and decide which one is correct. Use calculations and sketches to help explain your reasoning.
Writing an equation of a shifted parabola
The parabola is shifted horizontally:
-
If the parabola is shifted \(m\) units to the right, \(x\) is replaced by \((x-m)\).
-
If the parabola is shifted \(m\) units to the left, \(x\) is replaced by \((x+m)\).
The parabola is shifted vertically:
-
If the parabola is shifted \(n\) units down, \(y\) is replaced by \((y+n)\).
-
If the parabola is shifted \(n\) units up, \(y\) is replaced by \((y-n)\).
Worked example 6: Shifting a parabola
Given \(y = x^2 - 2x -3\).
-
If the parabola is shifted \(\text{1}\) unit to the right, determine the new equation of the parabola.
-
If the parabola is shifted \(\text{3}\) units down, determine the new equation of the parabola.
Determine the new equation of the shifted parabola
-
The parabola is shifted \(\text{1}\) unit to the right, so \(x\) must be replaced by \((x-1)\). \begin{align*} y &= x^2 - 2x -3\\ &= (x-1)^2 - 2(x-1) -3 \\ &= x^2 - 2x + 1 -2x + 2 - 3\\ &= x^2 - 4x \end{align*} Be careful not to make a common error: replacing \(x\) with \(x+1\) for a shift to the right.
-
The parabola is shifted \(\text{3}\) units down, so \(y\) must be replaced by \((y+3)\). \begin{align*} y + 3&= x^2 - 2x -3\\ y &= x^2 - 2x -3 - 3 \\ &= x^2 - 2x -6 \end{align*}
Sketching parabolas
Textbook Exercise 5.6
Sketch graphs of the following functions and determine:
- intercepts
- turning point
- axes of symmetry
- domain and range
\(y = -x^2 + 4x + 5\)
\(y = 2(x + 1)^2\)
\(y = 3x^2 - 2(x+2)\)
\(y = 3(x -2)^2 + 1\)
Draw the following graphs on the same system of axes:
\(f(x) = -x^2 + 7\)
\(g(x) = -(x - 2)^2 + 7\)
\(h(x) = (x - 2)^2 - 7\)
Draw a sketch of each of the following graphs:
\(y = ax^2 + bx + c\) if \(a > 0\), \(b > 0\), \(c < 0\).
\(y = ax^2 + bx + c\) if \(a < 0\), \(b = 0\), \(c > 0\).
\(y = ax^2 + bx + c\) if \(a < 0\), \(b < 0\), \(b^2 - 4ac < 0\).
\(y = (x+p)^2 + q\) if \(p < 0\), \(q < 0\) and the \(x\)-intercepts have different signs.
\(y = a(x+p)^2 + q\) if \(a < 0\), \(p < 0\), \(q > 0\) and one root is zero.
\(y = a(x+p)^2 + q\) if \(a > 0\), \(p = 0\), \(b^2 - 4ac > 0\).
Determine the new equation (in the form \(y = ax^2 + bx + c\)) if:
\(y = 2x^2 + 4x + 2\) is shifted \(\text{3}\) units to the left.
\begin{align*} y &=2x^2 + 4x + 2 \\ x &\Rightarrow x+3 \\ y_{\text{shifted}} &=2(x+3)^2 + 4(x+3) + 2 \\ &= 2(x^2 + 6x + 9) + 4x+12 + 2 \\ &= 2x^2 + 12x + 18 + 4x+12 + 2 \\ &= 2x^2 + 16x + 32 \end{align*}
\(y = -(x+1)^2\) is shifted \(\text{1}\) unit up.
\begin{align*} y &=2x^2 + 4x + 2 \\ y &\Rightarrow y-1 \\ y_{\text{shifted}} &= -(x+1)^2 + 1 \\ &= -(x^2 + 2x + 1) + 1 \\ &= -x^2 - 2x - 1 + 1 \\ &= -x^2 - 2x \end{align*}
\(y = 3(x-1)^2 + 2\left(x-\frac{1}{2}\right)\) is shifted \(\text{2}\) units to the right.
\begin{align*} y &= 3(x-1)^2 + 2\left(x-\frac{1}{2}\right) \\ x &\Rightarrow x-2 \\ y_{\text{shifted}} &= 3(x - 2-1)^2 + 2\left(x - 2 -\frac{1}{2}\right) \\ &= 3(x - 3)^2 + 2 \left(x - \frac{5}{2}\right) \\ &= 3(x^2 - 6x + 9) + 2x - 5 \\ &= 3x^2 - 18x + 27 + 2x - 5 \\ &= 3x^2 - 16x + 22 \end{align*}
Finding the equation of a parabola from the graph
If the intercepts are given, use \(y = a(x - x_1)(x - x_2)\).
| Example:  | \(x\)-intercepts: \((-1;0)\) and \((4;0)\) \(\begin{aligned} y &= a(x -x_1)(x - x_2) \\ &= a(x + 1)(x - 4) \\ &= ax^2 - 3ax - 4a \end{aligned}\)\(y\)-intercept: \((0;2)\) \(\begin{aligned} -4a &= 2 \\ a &= -\frac{1}{2} \end{aligned}\)Equation of the parabola: \(\begin{aligned} y &= ax^2 - 3ax - 4a \\ &= -\frac{1}{2}x^2 - 3\left(-\frac{1}{2}\right)x - 4\left(-\frac{1}{2}\right) \\ &= -\frac{1}{2}x^2 + \frac{3}{2}x + 2 \end{aligned}\) |
If the \(x\)-intercepts and another point are given, use \(y = a(x -x_1)(x - x_2)\).
| Example:  | \(x\)-intercepts: \((1;0)\) and \((5;0)\) \(\begin{aligned} y &= a(x -x_1)(x - x_2) \\ &= a(x - 1)(x - 5) \\ &= ax^2 - 6ax + 5a \end{aligned}\)Substitute the point: \((-1;\text{12})\) \(\begin{aligned} \text{12} &= a(-1)^2 - 6a(-1) + 5a \\ \text{12} &= a + 6a + 5a \\ \text{12} &= 12a \\ 1 &= a \end{aligned}\)Equation of the parabola: \(\begin{aligned} y &= ax^2 - 6ax + 5a \\ &= x^2 - 6x + 5 \end{aligned}\) |
If the turning point and another point are given, use \(y = a(x + p)^2 + q\).
| Example:  | Turning point: \((-3;1)\) \(\begin{aligned} y &= a(x + p)^2 + q \\ &= a(x + 3)^2 + 1 \\ &= ax^2 + 6ax + 9a + 1 \end{aligned}\)Substitute the point: \((1;5)\) \(\begin{aligned} 5 &= a(1)^2 + 6a(1) + 9a + 1 \\ 4 &= 16a \\ \frac{1}{4} &= a \end{aligned}\)Equation of the parabola: \(y = \frac{1}{4}(x + 3)^2 + 1\) |
Finding the equation
Textbook Exercise 5.7
Determine the equations of the following graphs. Write your answers in the form \(y = a(x + p)^2 + q\).
\begin{align*} y &= a(x + p)^2 + q \\ \text{Subst. } & (-1;6) \\ y &=a(x+1)^2+6 \\ &=ax^2+2ax+a+6 \\ y-\text{int: } &= (0;3) \\ 3 &=0 +0 +a+6 \\ \therefore 3 &= a + 6 \\ a &= -3 \\ y=-3(x+1)^2+6 &\text{ or } y =-3x^2-6x+3 \end{align*}
\begin{align*} y &= a(x + p)^2 + q \\ y&=a(x-0)(x-5) \\ &=ax^2-5ax \\ \text{Subst. } & (-1;3) \\ y &=ax^2-5ax \\ 3 &=a(-1)^2-5a(-1) \\ 3 &= a+5a \\ 3 &= 6a \\ \therefore a &= \frac{1}{2} \\ \therefore y&=\frac{1}{2}x^2-\frac{5}{2}x \end{align*}
\begin{align*} y &= a(x + p)^2 + q \\ \text{Subst. } &= (-2;0) \\ y &= a(x + 2)^2 \\ &=ax^2+4ax+4a \\ \text{Subst. } & (1;6) \\ 6 &= a +4a +4a \\ 6 &=9a \\ \therefore a &=\frac{2}{3} \\ \therefore y &= \frac{2}{3}(x+2)^2 \end{align*}
\begin{align*} y &= ax^2+bx+c \\ y &=ax^2+bx+4 \\ \text{Subst. }(1;6): \qquad 6&=a+b+4 \ldots (1) \\ \text{Eqn. }(1) \times 5: \qquad 30 &=5a+5b+20 \ldots (3) \\ \text{Subst. }(1;6): \qquad -6&=25a+5b+4 \ldots (2) \\ (2) - (3) \quad -36&=20a-16 \\ 20a&=-20 \\ a &= -1 \\ b&=3 \\ c&= 4 \\ \therefore y&=-x^2+3x+4 \end{align*}
