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End of chapter exercises

End of chapter exercises

Textbook Exercise 5.33

Show that if \(a<0\), then the range of \(f(x)=a{\left(x+p\right)}^{2}+q\) is \(\left\{f(x):f(x)\in \left(-\infty ,q\right]\right\}\).

\begin{align*} (x + p)^2 \geq 0\\ a(x + p)^2 \leq 0 \\ a(x + p)^2 + q < q \\ \therefore f(x) < q \end{align*}

If \(\left(2;7\right)\) is the turning point of \(f(x)=-2{x}^{2}-4ax+k\), find the values of the constants \(a\) and \(k\).

\(a = -2; k= -1\)

The following graph is represented by the equation \(f(x)=a{x}^{2}+bx\). The coordinates of the turning point are \(\left(3;9\right)\). Show that \(a=-1\) and \(b=6\).

\begin{align*} f(x) & = ax^2 + bx \\ x &= -{b}{2a} \\ \therefore 3 &= -\frac{b}{2a}\\ 6a &= -b\\ \text{subst. } (3;9) \quad 9 &= a(3)^2 + b(3) \\ 9 &= 9a + 3b \\ 9 &= 9a + 3(-6a) \\ 9 &= 9a -18a \\ 9 &= -9a \\ \therefore -1 &= a \\ \therefore b &= 6 \end{align*}

Given: \(f(x)={x}^{2}-2x+3\). Give the equation of the new graph originating if:

the graph of \(f\) is moved three units to the left.

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

the \(x\)-axis is moved down three units.

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

A parabola with turning point \(\left(-1;-4\right)\) is shifted vertically by \(\text{4}\) units upwards. What are the coordinates of the turning point of the shifted parabola?


Plot the graph of the hyperbola defined by \(y=\frac{2}{x}\) for \(-4\le x\le 4\). Suppose the hyperbola is shifted \(\text{3}\) units to the right and \(\text{1}\) unit down. What is the new equation then?


Based on the graph of \(y=\frac{k}{(x+p)} + q\), determine the equation of the graph with asymptotes \(y=2\) and \(x=1\) and passing through the point \(\left(2;3\right)\).

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

The columns in the table below give the \(y\)-values for the following functions: \(y={a}^{x}\), \(y={a}^{x+1}\) and \(y={a}^{x}+1\). Match each function to the correct column.

\(x\) A B C
\(-\text{2}\) \(\text{7,25}\) \(\text{6,25}\) \(\text{2,5}\)
\(-\text{1}\) \(\text{3,5}\) \(\text{2,5}\) \(\text{1}\)
\(\text{0}\) \(\text{2}\) \(\text{1}\) \(\text{0,4}\)
\(\text{1}\) \(\text{1,4}\) \(\text{0,4}\) \(\text{0,16}\)
\(\text{2}\) \(\text{1,16}\) \(\text{0,16}\) \(\text{0,064}\)

Column A: \(y={a}^{x} + 1\); Column B: \(y={a}^{x}\); Column C: \(y={a}^{x + 1}\)

The graph of \(f(x) = 1 + a \cdot 2^x\) (\(a\) is a constant) passes through the origin.

Determine the value of \(a\).

\begin{align*} 0 &= a \times 2^0 + 1 \\ \therefore -1 &= a \end{align*}

Determine the value of \(f(-15)\) correct to five decimal places.

\begin{align*} f(-15) &= -2^{-15} + 1 \\ &= \text{0,99997} \end{align*}

Determine the value of \(x\), if \(P\left(x;\text{0,5}\right)\) lies on the graph of \(f\).

\begin{align*} \text{0,5} &= -2^{x} + 1 \\ \text{0,5} &= 2^{x} \\ \therefore x &= -1 \end{align*}

If the graph of \(f\) is shifted \(\text{2}\) units to the right to give the function \(h\), write down the equation of \(h\).

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

The graph of \(f(x) = a \cdot b^x\) (\(a ≠ 0\)) has the point \(P\left(2;\text{144}\right)\) on \(f\).

If \(b = \text{0,75}\), calculate the value of \(a\).

\(a = 256\)

Hence write down the equation of \(f\).

\(f(x) = 256 \left( \frac{3}{4} \right)^x\)

Determine, correct to two decimal places, the value of \(f(13)\).

\(f(13) = \text{6,08}\)

Describe the transformation of the curve of \(f\) to \(h\) if \(h(x)=f(-x)\).

Shifted \(\text{2}\) units to the right

Using your knowledge of the effects of \(p\) and \(k\) draw a rough sketch of the following graphs without a table of values.

\(y = \sin 3\theta\) for \(-\text{180}\text{°} \leq \theta \leq \text{180}\text{°}\)


\(y = -\cos 2\theta\) for \(\text{0}\text{°} \leq \theta \leq \text{180}\text{°}\)


\(y = \tan \frac{1}{2}\theta\) for \(\text{0}\text{°} \leq \theta \leq \text{360}\text{°}\)


\(y = \sin (\theta-\text{45}\text{°})\) for \(-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}\)


\(y = \cos (\theta+\text{45}\text{°})\) for \(\text{0}\text{°} \leq \theta \leq \text{360}\text{°}\)


\(y = \tan (\theta-\text{45}\text{°})\) for \(\text{0}\text{°} \leq \theta \leq \text{360}\text{°}\)


\(y = 2\sin 2\theta\) for \(-\text{180}\text{°} \leq \theta \leq \text{180}\text{°}\)


\(y = \sin (\theta+\text{30}\text{°}) + 1\) for \(-\text{360}\text{°} \leq \theta \leq \text{0}\text{°}\)