Interpretation of Graphs

Use the sketch below to determine the values of \(a\) and \(q\) for the parabola of the form \(y=a{x}^{2}+q\).

Examine the sketch

From the sketch we see that the shape of the graph is a “frown”, therefore \(a<0\). We also see that the graph has been shifted vertically upwards, therefore \(q>0\).

Determine \(q\) using the \(y\)-intercept

The \(y\)-intercept is the point \((0;1)\).

\begin{align*} y& = a{x}^{2}+q \\ 1& = a{(0)}^{2}+q \\ \therefore q& = 1 \end{align*}

Use the other given point to determine \(a\)

Substitute point \((-1;0)\) into the equation:

\begin{align*} y& = a{x}^{2}+q \\ 0& = a{(-1)}^{2}+1 \\ \therefore a& = -1 \end{align*}

Write the final answer

\(a=-1\) and \(q=1\), so the equation of the parabola is \(y=-{x}^{2}+1\).

Use the sketch below to determine the values of \(a\) and \(q\) for the hyperbola of the form \(y=\cfrac{a}{x}+q\).

Examine the sketch

The two curves of the hyperbola lie in the second and fourth quadrant, therefore \(a<0\). We also see that the graph has been shifted vertically upwards, therefore \(q>0\).

Substitute the given points into the equation and solve

Substitute the point \((-1;2)\):

\begin{align*} y& = \cfrac{a}{x}+q \\ 2& = \cfrac{a}{-1}+q \\ \therefore 2& = -a+q \end{align*}

Substitute the point \((1;0)\):

\begin{align*} y& = \cfrac{a}{x}+q \\ 0& = \cfrac{a}{1}+q \\ \therefore a& = -q \end{align*}

Solve the equations simultaneously using substitution

\begin{align*} 2& = -a+q \\ & = q+q \\ & = 2q \\ \therefore q& = 1 \\ \therefore a& = -q \\ & = -1 \end{align*}

Write the final answer

\(a=-1\) and \(q=1\), therefore the equation of the hyperbola is \(y=\dfrac{-1}{x}+1\).

The graphs of \(y=-{x}^{2}+4\) and \(y=x-2\) are given. Calculate the following:

1. coordinates of \(A\), \(B\), \(C\), \(D\)

2. coordinates of \(E\)

3. distance \(CD\)

Calculate the intercepts

For the parabola, to calculate the \(y\)-intercept, let \(x=0\):

\begin{align*} y& = -{x}^{2}+4 \\ & = -{0}^{2}+4 \\ & = 4 \end{align*}

This gives the point \(C(0;4)\).

To calculate the \(x\)-intercept, let \(y=0\):

\begin{align*} y& = -{x}^{2}+4 \\ 0& = -{x}^{2}+4 \\ {x}^{2}-4& = 0 \\ (x+2)(x-2)& = 0 \\ \therefore x& = ±2 \end{align*}

This gives the points \(A(-2;0)\) and \(B(2;0)\).

For the straight line, to calculate the \(y\)-intercept, let \(x=0\):

\begin{align*} y& = x-2 \\ & = 0-2 \\ & = -2 \end{align*}

This gives the point \(D(0;-2)\).

Calculate the point of intersection \(E\)

At \(E\) the two graphs intersect so we can equate the two expressions:

\begin{align*} x-2& = -{x}^{2}+4 \\ \therefore {x}^{2}+x-6& = 0 \\ \therefore (x-2)(x+3)& = 0 \\ \therefore x& = 2 \text{ or } -3 \end{align*}

At \(E\), \(x=-3\), therefore \(y=x-2=-3-2=-5\). This gives the point \(E(-3;-5)\).

Calculate distance \(CD\)

\begin{align*} CD& = CO+OD \\ & = 4+2 \\ & = 6 \end{align*}

Distance \(CD\) is \(\text{6}\) units.

Write the final answer

1. coordinates of \(A(-2;0)\), \(B(2;0)\), \(C(0;4)\), \(D(0;-2)\)

2. coordinates of \(E(-3;-5)\)

3. distance \(CD = 6\) units

Use the sketch to determine the equation of the trigonometric function \(f\) of the form \(y=a f(\theta)+q\).

Examine the sketch

From the sketch we see that the graph is a sine graph that has been shifted vertically upwards. The general form of the equation is \(y=a\sin\theta +q\).

Substitute the given points into equation and solve

At \(N\), \(\theta =210°\) and \(y=0\):

\begin{align*} y& = a\sin\theta +q \\ 0& = a\sin 210°+q \\ & = a(-\cfrac{1}{2})+q \\ \therefore q& = \cfrac{a}{2} \end{align*}

At \(M\), \(\theta = 90°\) and \(y=\cfrac{3}{2}\):

\begin{align*} \cfrac{3}{2}& = a\sin90°+q \\ & = a+q \end{align*}

Solve the equations simultaneously using substitution

\begin{align*} \cfrac{3}{2}& = a+q \\ & = a+\cfrac{a}{2} \\ 3& = 2a+a \\ 3a& = 3 \\ \therefore a& = 1 \\ \therefore q& = \cfrac{a}{2} \\ & = \cfrac{1}{2} \end{align*}