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Exponential Functions

5.4 Exponential functions (EMBGS)

Revision (EMBGT)

Functions of the form \(y=a{b}^{x}+q\)

Functions of the general form \(y=a{b}^{x}+q\), for \(b>0\), are called exponential functions, where \(a\), \(b\) and \(q\) are constants.

The effects of \(a\), \(b\) and \(q\) on \(f(x) = ab^x + q\):

  • The effect of \(q\) on vertical shift

    • For \(q>0\), \(f(x)\) is shifted vertically upwards by \(q\) units.

    • For \(q<0\), \(f(x)\) is shifted vertically downwards by \(q\) units.

    • The horizontal asymptote is the line \(y = q\).

  • The effect of \(a\) on shape

    • For \(a>0\), \(f(x)\) is increasing.

    • For \(a<0\), \(f(x)\) is decreasing. The graph is reflected about the horizontal asymptote.

  • The effect of \(b\) on direction

    Assuming \(a > 0\):

    • If \(b > 1\), \(f(x)\) is an increasing function.
    • If \(0 < b < 1\), \(f(x)\) is a decreasing function.
    • If \(b \leq 0\), \(f(x)\) is not defined.

\(b>1\)

\(a<0\)

\(a>0\)

\(q>0\)

9e5ebcaf6bcbc05ffef94cb02e120f14.pnga0202195ebbe945b0b3a1db523dbddba.png

\(q<0\)

bc406680884a94205fb4c3249bb77dd9.pngdeed22526c559254713ae52f0b53c858.png

\(0<b<1\)

\(a<0\)

\(a>0\)

\(q>0\)

a0e5607c40becee4cb0e0db3ad6896f0.png4fe3dbe184c6823ca683d66a10cd6e60.png

\(q<0\)

cb69198bd07e3b8aea3bbde878f01f3b.png072cdac28a3cab0af8505f60cda2cd59.png

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Revision

Exercise 5.15

On separate axes, accurately draw each of the following functions:

  • Use tables of values if necessary.
  • Use graph paper if available.

\(y_1 = 3^x\)

4f24822a8760541af73bb498b5d3d571.png

\(y_2 = -2 \times 3^x\)

22870e4663a2518f7717b0b31f59bef6.png

\(y_3 = 2 \times 3^x + 1\)

ae645988f28c6c8b855c70a91b9919ec.png

\(y_4 = 3^x - 2\)

7db0d02a3587bc695202706fcc1926d7.png

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 \(q\)no vertical shift
value of \(a\)\(a = 1\)
effect of \(a\)increasing
asymptote

\(x\)-axis, \(y = 0\)

domain\(\{x: x \in \mathbb{R} \}\)
range\(\{y: y \in \mathbb{R}, y > 0 \}\)
\(y_1\)\(y_2\)\(y_3\)\(y_4\)
value of \(q\)\(q = 0\)\(q = 0\)\(q = 1\)\(q = 2\)
effect of \(q\)no vertical shiftno vertical shiftshift \(1\) unit upshift \(2\) units down
value of \(a\)\(a = 1\)\(a = -2\)\(a = 2\)\(a = 1\)
effect of \(a\)increasingdecreasingincreasingincreasing
asymptote

\(x\)-axis, \(y = 0\)

\(x\)-axis, \(y = 0\)

\(y = 1\)

\(y = -2\)

domain\(\{x: x \in \mathbb{R} \}\)\(\{x: x \in \mathbb{R} \}\)\(\{x: x \in \mathbb{R} \}\)\(\{x: x \in \mathbb{R} \}\)
range\(\{y: y \in \mathbb{R}, y > 0 \}\)\(\{y: y \in \mathbb{R}, y < 0 \}\)\(\{y: y \in \mathbb{R}, y > 1 \}\)\(\{y: y \in \mathbb{R}, y > -2 \}\)

Functions of the form \(y=a{b}^{\left(x+p\right)}+q\) (EMBGV)

We now consider exponential functions of the form \(y=a{b}^{\left(x+p\right)}+q\) and the effects of parameter \(p\).

The effects of \(a\), \(p\) and \(q\) on an exponential graph

  1. On the same system of axes, plot the following graphs:

    1. \(y_1 = 2^x\)
    2. \(y_2 = 2^{(x - 2)}\)
    3. \(y_3 = 2^{(x - 1)}\)
    4. \(y_4 = 2^{(x + 1)}\)
    5. \(y_5 = 2^{(x + 2)}\)

    Use your sketches of the functions above to complete the following table:

    \(y_1\)\(y_2\)\(y_3\)\(y_4\)\(y_5\)
    intercept(s)
    asymptote
    domain
    range
    effect of \(p\)
  2. On the same system of axes, plot the following graphs:

    1. \(y_1 = 2^{(x - 1)} + 2\)
    2. \(y_2 = 3 \times 2^{(x - 1)} + 2\)
    3. \(y_3 = \frac{1}{2} \times 2^{(x - 1)} + 2\)
    4. \(y_4 = 0 \times 2^{(x - 1)} + 2\)
    5. \(y_5 = -3 \times 2^{(x - 1)} + 2\)

    Use your sketches of the functions above to complete the following table:

    \(y_1\)\(y_2\)\(y_3\)\(y_4\)\(y_5\)
    intercept(s)
    asymptotes
    domain
    range
    effect of \(a\)

The effect of the parameters on \(y = ab^{x + p} + q\)

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 effect of \(q\) is a vertical shift. The value of \(q\) also affects the horizontal asymptotes, the line \(y = q\).

The value of \(a\) affects the shape of the graph and its position relative to the horizontal asymptote.

  • For \(a>0\), the graph lies above the horizontal asymptote, \(y = q\).

  • For \(a<0\), the graph lies below the horizontal asymptote, \(y = q\).

\(p>0\)\(p<0\)
\(a<0\)\(a>0\)\(a<0\)\(a>0\)
\(q>0\)2599045f9ae5f243b8a56b9fb0059ae1.pnge4f1c672d1ba75f1a08dc1dd8a18c11b.png2165531844925fb1ac73d334e84b9b19.png2abea4bd1bbe65c4a2314111f669823a.png

\(q<0\)

58507007c3b745f66456fe3f5eb7ab99.pngc72d49ca236764c97bc089770b02ccf5.png99a4378764e447359f56f477a4950fc4.pngb194ad64d87973241711e1839810a406.png

Discovering the characteristics

For functions of the general form: \(f(x) = y = ab^{(x+p)} + q\):

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{align*} {b}^{\left(x+p\right)} & > 0 \\ a {b}^{\left(x+p\right)} & > 0 \\ a {b}^{\left(x+p\right)} + q & > q \\ f(x) & > q \end{align*} The range is therefore \(\{ y: y > q, y \in \mathbb{R} \}\).

Similarly, if \(a < 0\), the range is \(\{ y: y < q, y \in \mathbb{R} \}\).

Worked example 14: Domain and range

State the domain and range for \(g(x) = 5 \times 3^{(x+1)} - 1\).

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*} 3^{(x+1)} & > 0\\ 5 \times 3^{(x+1)} & > 0\\ 5 \times 3^{(x+1)} - 1 & > -1\\ \therefore g(x) & > -1 \end{align*} Therefore the range is \(\{g(x): g(x) > -1 \}\) or in interval notation \((-1; \infty)\).

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Domain and range

Exercise 5.16

Give the domain and range for each of the following functions:

\(y = \left( \frac{3}{2} \right)^{(x + 3)}\)

\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y > 0, y\in \mathbb{R} \right \} \end{align*}

\(f(x) = -5^{(x - 2)} + 1\)

\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y < 1, y\in \mathbb{R} \right \} \end{align*}

\(y + 3 = 2^{(x + 1)}\)

\begin{align*} y + 3 &= 2^{(x + 1)} \\ y &= 2^{(x + 1)} - 3 \\ \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y > -3, y\in \mathbb{R} \right \} \end{align*}

\(y = n + 3^{(x - m)}\)

\begin{align*} y &= n + 3^{(x - m)} \\ y &= 3^{(x - m)} + n \\ \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y > n, y\in \mathbb{R} \right \} \end{align*}

\(\frac{y}{2} = 3^{(x - 1)} - 1\)

\begin{align*} \frac{y}{2} &= 3^{(x - 1)} - 1 \\ y &= 2 \times 3^{(x - 1)} - 2 \\ \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y > 2, y\in \mathbb{R} \right \} \end{align*}

Intercepts

The \(y\)-intercept:

To calculate the \(y\)-intercept we let \(x=0\). For example, the \(y\)-intercept of \(g(x) = 3 \times 2^{(x + 1)} + 2\) is determined by setting \(x=0\): \begin{align*} g(0) &= 3 \times 2^{(0 + 1)} + 2 \\ &= 3 \times 2 + 2\\ &= 8 \end{align*} This gives the point \((0;8)\).

The \(x\)-intercept:

To calculate the \(x\)-intercept we let \(y=0\). For example, the \(x\)-intercept of \(g(x) = 3 \times 2^{(x + 1)} + 2\) is determined by setting \(y=0\): \begin{align*} 0 &= 3 \times 2^{(x + 1)} + 2 \\ -2 &= 3 \times 2^{(x + 1)} \\ -\frac{2}{3} &= 2^{(x + 1)} \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

Exercise 5.17

Determine the \(x\)- and \(y\)-intercepts for each of the following functions:

\(f(x) = 2^{(x + 1)} - 8\)

\begin{align*} \text{For } x=0 \quad y &= 2^{(0 + 1)} - 8 \\ &= 2 - 8 \\ &= -6 \\ \therefore & (0;-6) \\ \text{For } y=0 \quad 0 &= 2^{(x + 1)} - 8 \\ 2^3 &= 2^{(x + 1)} \\ \therefore 3 &= x + 1 \\ \therefore 2 &= x \\ \therefore & (2;0) \end{align*}

\(y = 2 \times 3^{(x - 1)} - \text{18}\)

\begin{align*} \text{For } x=0 \quad y &= 2 \times 3^{(0 - 1)} - 18 \\ &= \frac{2}{3} -18 \\ &= -17\frac{1}{3} \\ \therefore & (0;-17\frac{1}{3}) \\ \text{For } y=0 \quad 0 &= 2 \times 3^{(x - 1)} - 18 \\ 18 &= 2 \times 3^{(x - 1)} \\ 9 &= 3^{(x - 1)} \\ 3^2 &= 3^{(x - 1)} \\ \therefore 2 &= x - 1 \\ \therefore 3 &= x \\ \therefore & (3;0) \end{align*}

\(y + 5^{(x + 2)} = 5\)

\begin{align*} y + 5^{(x + 2)} &= 5 \\ y &= -5^{(x + 2)} + 5 \\ \text{For } x=0 \quad y &= -5^{(0 + 2)} + 5 \\ &= -25 + 5 \\ &= -20 \\ \therefore & (0;-20) \\ \text{For } y=0 \quad 0 &= -5^{(x + 2)} + 5 \\ 5^{(x + 2)} &= 5\\ \therefore x + 2 &= 1 \\ \therefore x &= -1 \\ \therefore & (-1;0) \end{align*}

\(y = \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} - \text{0,75}\)

\begin{align*} y &= \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} - \text{0,75} \\ y &= \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} - \frac{3}{4} \\ \text{For } x=0 \quad y &= \frac{1}{2} \left( \frac{3}{2} \right)^{(0 + 3)} - \frac{3}{4} \\ &= \frac{1}{2} \left( \frac{3}{2} \right)^{3} - \frac{3}{4} \\ &= \frac{1}{2} \left( \frac{27}{8} \right)- \frac{3}{4} \\ &= \frac{27}{16} - \frac{3}{4} \\ &= \frac{15}{16} \\ \therefore & (0;\frac{15}{16}) \\ \text{For } y=0 \quad 0 &= \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} - \frac{3}{4} \\ \frac{3}{4} &= \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} \\ \frac{3}{2} &= \left( \frac{3}{2} \right)^{(x + 3)} \\ \therefore 1 &= x + 3 \\ \therefore -2 &= x \\ \therefore & (-2;0) \end{align*}

Asymptote

Exponential functions of the form \(y = ab^{(x+p)} + q\) have a horizontal asymptote, the line \(y = q\).

Worked example 15: Asymptote

Determine the asymptote for \(y = 5 \times 3^{(x+1)} - 1\).

Determine the asymptote

The asymptote of \(g(x)\) can be calculated as: \begin{align*} 3^{(x+1)} & \ne 0\\ 5 \times 3^{(x+1)} & \ne 0\\ 5 \times 3^{(x+1)} - 1 & \ne -1\\ \therefore y & \ne -1 \end{align*} Therefore the asymptote is the line \(y = -1\).

Asymptote

Exercise 5.18

Give the asymptote for each of the following functions:

\(y = -5^{(x + 1)}\)

\begin{align*} y &= -5^{(x + 1)} \\ \text{Horizontal asymptote: } \quad y &= 0 \end{align*}

\(y = 3^{(x - 2)} + 1\)

\begin{align*} y &= 3^{(x - 2)} + 1 \\ \text{Horizontal asymptote: } \quad y &= 1 \end{align*}

\(\left( \frac{3y}{2} \right) = 5^{(x + 3)} - 1\)

\begin{align*} \left( \frac{3y}{2} \right) &= 5^{(x + 3)} - 1 \\ 3y &= 2 \times 5^{(x + 3)} - 2 \\ y &= \frac{2}{3} \times 5^{(x + 3)} - \frac{2}{3} \\ \text{Horizontal asymptote: } \quad y &= -\frac{2}{3} \end{align*}

\(y = 7^{(x + 1)} - 2\)

\begin{align*} y &= 7^{(x + 1)} - 2 \\ \text{Horizontal asymptote: } \quad y &= -2 \end{align*}

\(\frac{y}{2} + 1 = 3^{(x + 2)}\)

\begin{align*} \frac{y}{2} + 1 &= 3^{(x + 2)} \\ \frac{y}{2} &= 3^{(x + 2)} - 1 \\ y &= 2 \times 3^{(x + 2)} - 2 \\ \text{Horizontal asymptote: } \quad y &= -2 \end{align*}

Sketching graphs of the form \(f(x)=a{b}^{\left(x+p\right)}+q\)

In order to sketch graphs of functions of the form, \(f(x)=a{b}^{\left(x+p\right)}+q\), we need to determine five characteristics:

  • shape

  • \(y\)-intercept

  • \(x\)-intercept

  • asymptote

  • domain and range

Worked example 16: Sketching an exponential graph

Sketch the graph of \(2y = \text{10} \times 2^{(x+1)} - 5\).

Mark the intercept(s) and asymptote. State the domain and range of the function.

Examine the equation of the form \(y = ab^{(x + p)} + q\)

We notice that \(a > 0\) and \(b > 1\), therefore the function is increasing.

Determine the \(y\)-intercept

The \(y\)-intercept is obtained by letting \(x = 0\): \begin{align*} 2y &= \text{10} \times 2^{(0+1)} - 5\\ &= \text{10} \times 2 - 5\\ &= \text{15}\\ \therefore y &= 7\frac{1}{2} \end{align*} This gives the point \((0;7\frac{1}{2})\).

Determine the \(x\)-intercept

The \(x\)-intercept is obtained by letting \(y = 0\): \begin{align*} 0 &= \text{10} \times 2^{(x+1)} - 5\\ 5 &= \text{10} \times 2^{(x+1)} \\ \frac{1}{2} &= 2^{(x+1)}\\ 2^{-1} &= 2^{(x+1)}\\ \therefore -1 &= x + 1 \quad \text{(same base)}\\ -2 &= x \end{align*} This gives the point \((-2;0)\).

Determine the asymptote

The horizontal asymptote is the line \(y = -\frac{5}{2}\).

Plot the points and sketch the graph

cedd30d9cc9b8f7fd1c633be14f857a7.png

State the domain and range

Domain: \(\{ x: x \in \mathbb{R} \}\)

Range: \(\{ y: y > -\frac{5}{2}, y \in \mathbb{R} \}\)

Finding the equation of an exponential function from the graph

Worked example 17: Finding the equation of an exponential function from the graph

Use the given graph of \(y = -2 \times 3^{(x + p)} + q\) to determine the values of \(p\) and \(q\).

34167a123433bd5bc809018e0c77db9c.png

Examine the equation of the form \(y = ab^{(x + p)} + q\)

From the graph we see that the function is decreasing. We also note that \(a = -2\) and \(b = 3\).

We need to solve for \(p\) and \(q\).

Use the asymptote to determine \(q\)

The horizontal asymptote \(y = 6\) is given, therefore we know that \(q = 6\). \[y = -2 \times 3^{(x + p)} + 6\]

Use the \(x\)-intercept to determine \(p\)

Substitute \((2;0)\) into the equation and solve for \(p\): \begin{align*} y &= -2 \times 3^{(x + p)} + 6 \\ 0 &= -2 \times 3^{(2 + p)} + 6 \\ -6 &= -2 \times 3^{(2 + p)} \\ 3 &= 3^{(2 + p)} \\ \therefore 1 &= 2 + p \quad \text{(same base)}\\ \therefore p &= -1 \end{align*}

Write the final answer

\[y = -2 \times 3^{(x - 1)} + 6\]

Mixed exercises

Exercise 5.19

Given the graph of the hyperbola of the form \(h(x) = \frac{k}{x}\), \(x < 0\), which passes though the point \(A(-\frac{1}{2}; -6)\).

721cee6f733bcb87b121912982419919.png

Show that \(k=3\).

\begin{align*} y &=\frac{k}{x} \\ \text{Subst. } (-\frac{1}{2}; -6) \qquad -6 &=\frac{k}{-\frac{1}{2}} \\ -6 \times -\frac{1}{2} &= k \\ \therefore k &=3 \\ \therefore h(x) &= \frac{3}{x} \end{align*}

Write down the equation for the new function formed if \(h(x)\):

is shifted \(\text{3}\) units vertically upwards

\begin{align*} h(x) &= \frac{3}{x} \\ \therefore y &\Rightarrow y - 3 \\ y - 3 &=\frac{3}{x} \\ y &=\frac{3}{x}+3 \end{align*}

is shifted to the right by \(\text{3}\) units

\begin{align*} h(x) &= \frac{3}{x} \\ \therefore x &\Rightarrow x - 3 \\ y &=\frac{3}{x -3} \end{align*}

is reflected about the \(y\)-axis

\begin{align*} h(x) &= \frac{3}{x} \\ \therefore x &\Rightarrow -x \\ y &=\frac{3}{x -3} \end{align*}

is shifted so that the asymptotes are \(x = 0\) and \(y = -\frac{1}{4}\)

\begin{align*} h(x) &= \frac{3}{x} \\ \therefore p=0 &\text{ and } q = -\frac{1}{4} \\ y &=\frac{3}{x} -\frac{1}{4} \end{align*}

is shifted upwards to pass through the point \((-1;1)\)

\begin{align*} h(x) &= \frac{3}{x} \\ \therefore y &\Rightarrow y + m \\ y &=\frac{3}{x} + m \\ \text{Subst.} (-1;1) \qquad 1 &= \frac{3}{-1} + m \\ 1 + 3 &= + m \\ \therefore m &= 4 \\ \therefore y &=\frac{3}{x} + 4 \end{align*}

is shifted to the left by \(\text{2}\) units and \(\text{1}\) unit vertically downwards (for \(x < 0\))

\begin{align*} h(x) &= \frac{3}{x} \\ \therefore x &\Rightarrow x + 2 \\ \therefore y &\Rightarrow y + 1 \\ y + 1 &=\frac{3}{x + 2} \\ \therefore y &=\frac{3}{x + 2} - 1 \end{align*}

Given the graphs of \(f(x) = a(x+p)^2\) and \(g(x) = \frac{a}{x}\).

The axis of symmetry for \(f(x)\) is \(x = -1\) and \(f(x)\) and \(g(x)\) intersect at point \(M\). The line \(y = 2\) also passes through \(M\).

6669aaf1e22ad0e9ccc69674c9d8fcd9.png

Determine:

the coordinates of \(M\)

\(f(x)\) is symmetrical about the line \(x = -1\), therefore \(M(-2;2)\).

the equation of \(g(x)\)

\begin{align*} g(x) &= \frac{a}{x} \\ \text{Subst. } M(-2;2) \qquad 2 &= \frac{a}{-2} \\ \therefore a &= -4 \\ \therefore g(x)&=\frac{-4}{x} \end{align*}

the equation of \(f(x)\)

\begin{align*} f(x) &= a(x + p)^2 + q \\ \text{No vertical shift } \therefore q &= 0 \\ \text{Axis of symmetry } x = -1 \qquad \therefore f(x) &= a(x + 1)^2 \\ \text{Subst. } M(-2;2) \qquad 2 &= a(-2 + 1)^2 \\ 2 &= a(-1)^2 \\ \therefore a &= 2 \\ \therefore f(x) &= 2(x + 1)^2 \end{align*}

the values for which \(f(x) < g(x)\)

\(- 2 < x < 0\)

the range of \(f(x)\)

\(\text{Range: }\left \{ y: y\in \mathbb{R}, y \geq 0 \right \}\)

On the same system of axes, sketch:

the graphs of \(k(x) = 2(x + \frac{1}{2})^2 - 4\frac{1}{2}\) and \(h(x) = 2^{(x + \frac{1}{2})}\). Determine all intercepts, turning point(s) and asymptotes.

e7319a9fcb3b0ac4826dbc899907a2cb.png

the reflection of \(h(x)\) about the \(x\)-axis. Label this function as \(j(x)\).

55f80f44e6741930a08f5cf9c6014483.png

Sketch the graph of \(y = ax^2 + bx + c\) for:

\(a < 0\), \(b > 0\), \(b^2 < 4ac\)

71c4f5d1db0ca7873172e0ff14b329e9.png

\(a > 0\), \(b > 0\), one root \(=0\)

f032edc8df27fb81cbc626a01cf97a4a.png

On separate systems of axes, sketch the graphs:

\(y = \frac{2}{x - 2}\)

\(y = \frac{2}{x} - 2\)

\(y = -2^{(x - 2)}\)

8ca24315e1079d186b90ed14632fb7a1.png

For the diagrams shown below, determine:

  • the equations of the functions; \(f(x) = a(x + p)^2 + q\), \(g(x)=ax^2 + q\), \(h(x) = \frac{a}{x}, x < 0\) and \(k(x)=b^x + q\)
  • the axes of symmetry of each function
  • the domain and range of each function
e6870179c461368951616b8a6423eecb.png
\begin{align*} f(x) &= a(x + p)^2 + q \\ \text{From turning point: } p= -2 &\text{ and } q = 3 \\ \therefore f(x) &= a(x - 2 )^2 + 3 \\ \text{Subst. } (0;0) \qquad 0 &= a(0 - 2)^2 + 3 \\ -3 &= 4a \\ \therefore a &= -\frac{3}{4} \\ f(x) &= -\frac{3}{4}(x - 2)^2 + 3 \\ \text{Axes of symmetry: } x &= 2 \\ \text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y:y \in \mathbb{R}, y \leq 3 \right \} \end{align*}
6cabdd83dd9c469266e1d9ec30bb9cf9.png
\begin{align*} g(x) &= ax^2 + q \\ \text{From turning point: } p= 0 &\text{ and } q = -2 \\ \therefore g(x) &= a(x)^2 - 2 \\ \text{Subst. } (-2;-1) \qquad -1 &= a(-2 )^2 - 2\\ -1 &= 4a - 2\\ 1 &= 4a - 2\\ \therefore a &= \frac{1}{4} \\ g(x) &= \frac{1}{4}x^2 -2 \\ \text{Axes of symmetry: } x &= 0 \\ \text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y:y \in \mathbb{R}, y \geq -2 \right \} \\ h(x) &= \frac{a}{x + p} + q \\ \text{From graph: } p= 0 &\text{ and } q = 0 \\ h(x) &= \frac{a}{x} \\ \text{Subst. } (-2;-1) \qquad -1 &=\frac{a}{-2} \\ 2 &= a \\ \therefore h(x) &= \frac{2}{x} \\ \text{Axes of symmetry: } y &= x \\ \text{Domain: } &\left \{ x:x \in \mathbb{R}, x < 0 \right \} \\ \text{Range: } & \left \{ y:y \in \mathbb{R}, y < 0 \right \} \end{align*}
f1cfc1463e3ed913bf3a4bac1be16a6b.png
\begin{align*} y &= 2^{x} + \frac{1}{2} \\ \text{Reflect about } x = 0 \qquad \therefore x &\\Rightarrow -x \\ \therefore k(x) &= 2^{-x} + \frac{1}{2} \\ &= \left( \frac{1}{2} \right)^{x} + \frac{1}{2} \\ \text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y:y \in \mathbb{R}, y > \frac{1}{2} \right \} \end{align*}

Given the graph of the function \(Q(x) = a^x\).

d31841e6df3c41df220f0a157c99b421.png

Show that \(a = \frac{1}{3}\).

\begin{align*} y&=a^x \\ \text{Subst.} \left( 1;\frac{1}{3} \right) \qquad \frac{1}{3} &=a^{1} \\ \therefore a &= \frac{1}{3} \end{align*}

Find the value of \(p\) if the point \((-2;p)\) is on \(Q\).

\begin{align*} Q(x) &= \left( \frac{1}{3} \right)^x \\ \text{Subst.} \left( -2;p \right) p &= \left( \frac{1}{3} \right)^{-2} \\ p &= 9 \end{align*}

Calculate the average gradient of the curve between \(x = -2\) and \(x = 1\).

\begin{align*} \text{Average gradient} &= \frac{\left ( \frac{1}{3} \right )^{-2} - \left ( \frac{1}{3} \right )^1}{-2-(1)} \\ &= \frac{9-\frac{1}{3}}{-3} \\ &=\frac{8\frac{2}{3}}{-3} \\ &= \frac{-26}{9} \\ &= -2\frac{8}{9} \end{align*}

Determine the equation of the new function formed if \(Q\) is shifted \(\text{2}\) units vertically downwards and \(\text{2}\) units to the left.

\begin{align*} Q(x) &= \left( \frac{1}{3} \right)^x \\ \therefore x & \Rightarrow x + 2 \\ \therefore y & \Rightarrow y + 2 \\ y + 2 &= \left( \frac{1}{3} \right)^{x + 2} \\ y &= \left( \frac{1}{3} \right)^{x + 2} -2 \end{align*}

Find the equation for each of the functions shown below:

\(f(x) = 2^x + q\)

\(g(x) = mx + c\)

49123fe8f8bc3d0683fec8c8296984d8.png
\begin{align*} f(x) &= 2^x + q \\ \text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= 2^{0} + q \\ -\frac{1}{2} &= 1 + q \\ \therefore q &= -\frac{3}{2} \\ \therefore f(x) &=2^x-\frac{3}{2} \\ g(x) &= mx +c \\ \text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= m(0) + c \\ \therefore c &= -\frac{1}{2} \\ g(x) &= mx -\frac{1}{2} \\ \text{Subst. } (-2;0) \quad 0 &= m(-2) -\frac{1}{2} \\ \frac{1}{2} &= -2m \\ \therefore m &= -\frac{1}{4} \\ \therefore g(x) &= -\frac{1}{4}x-\frac{1}{2} \end{align*}

\(h(x) = \frac{k}{x + p} + q\)

712744df743364c7d8d76f459772d1a9.png
\begin{align*} h(x) &= \frac{k}{x + p} + q \\ \text{From graph: } \quad p = -2 \\ h(x) &= \frac{k}{x + 2} + q \\ \text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= \frac{k}{2} + q \\ -1 &= k + 2q \ldots (1)\\ \text{Subst. } (1;0) \quad 0 &= \frac{k}{1 + 2} + q \\ 0 &= k + 3q \ldots (2)\\ (2) - (1): \qquad 1 &= 0 + q \\ \therefore q &= 1 \\ \text{and } k &= -3\\ \therefore h(x) &= -\frac{3}{x + 2} + 1 \end{align*}

Given: the graph of \(k(x) = -x^2 + 3x + \text{10}\) with turning point at \(D\). The graph of the straight line \(h(x) = mx + c\) passing through points \(B\) and \(C\) is also shown.

0cd6032ba1041663620ec1d84e2dd6a2.png

Determine:

the lengths \(AO\), \(OB\), \(OC\) and \(DE\)

\begin{align*} y &=-x^2+3x+10 \\ \text{Let } y &= 0 \\ 0 &=-x^2+3x+10 \\ &= x^2-3x-10\\ &= (x-5)(x+2)\\ \therefore x=5 &\text{ or }x =-2 \\ \therefore AO &= \text{2}\text{ units} \\ \therefore BO &=\text{5}\text{ units} \\ CO&=\text{10}\text{ units}\\ \text{Axes of symmetry: } x &=-\frac{5 - 2}{2} \\ &=\frac{3}{2} \\ \text{Subst. } x &= \frac{3}{2} \\ y &=-\left ( \frac{3}{2} \right )^2+3\frac{3}{2}+10 \\ &=12\frac{1}{4} \\ \therefore DE &= \text{12,25}\text{ units} \end{align*}

the equation of \(DE\)

\(DE = 12\frac{1}{4}\)

the equation of \(h(x)\)

\begin{align*} h(x) & = mx + c \\ h(x) & = mx + 10 \\ \text{Subst. } (5;0) \qquad 0 &= m(5) + 10 \\ -10 &= 5m \\ \therefore m &= -2 \\ \therefore h(x) &=-2x+10 \end{align*}

the \(x\)-values for which \(k(x) < 0\)

\(\left \{ x: x\in \mathbb{R}, x < -2 \text{ and } x > 5 \right \}\)

the \(x\)-values for which \(k(x) \geq h(x)\)

\(\left \{ x: x\in \mathbb{R}, 0 \leq x \leq 5 \right \}\)

the length of \(DF\)

\begin{align*} \text{At } x = \frac{3}{2} k(x) &= 12\frac{1}{4} \\ \text{At } x = \frac{3}{2} h(x) &= -2 \left( \frac{3}{2} \right) + 10 \\ &= -3 + 10 \\ &= 7\\ \therefore DF &= 12\frac{1}{4} - 7 \\ &= 12\frac{1}{4} - 7 \\ &= \text{5,25}\text{ units} \end{align*}

Trigonometric functions are examined in PAPER 2.