Mathematics Functions Grade 12 Pdf

2.1 Revision (EMCF6)

In previous grades we learned about the characteristics of linear, quadratic, hyperbolic and exponential functions. In this chapter we will demonstrate the ability to work with various types of functions and relations including inverses. In particular, we will look at the graphs of the inverses of:

Linear functions: \(y=mx+c\) or \(y = ax + q\)

Quadratic functions: \(y=a{x}^{2}\)

Exponential functions: \(y=b^{x} \quad (b > 0, b \ne 1)\)

Worked example 1: Linear function

Draw a graph of \(2y + x - 8 = 0\) and determine the significant characteristics of this linear function.

Write the equation in standard form \(y = mx + c\)

\begin{align*} 2y + x - 8 &= 0 \\ 2y &= - x + 8 \\ \therefore y &= -\frac{1}{2}x + 4 \\ \therefore m &= -\frac{1}{2} \\ \text{And } c &= 4 \end{align*}

Draw the straight line graph

To draw the straight line graph we can use the gradient-intercept method:

\begin{align*} y-\text{intercept}: \enspace & (0;4) \\ m &= -\frac{1}{2} \end{align*}

Alternative method: we can also determine and plot the \(x\)- and \(y\)-intercepts as follows:

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

\begin{align*} y &= -\frac{1}{2}(0) + 4 \\ \therefore y &= 0 + 4 \\ &= 4 \end{align*}

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

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

\begin{align*} 0 &= -\frac{1}{2}x + 4 \\ \frac{1}{2}x &= 4 \\ \therefore x &= 8 \end{align*}

This gives the point \((8;0)\).

Determine the characteristics

Gradient: \(-\frac{1}{2}\)

Intercepts: \((0;4) \text{ and } (8;0)\)

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

Range: \(\{y: y \in \mathbb{R} \}\)

Decreasing function: as \(x\) increases, \(y\) decreases.

Worked example 2: Quadratic function

Write the quadratic function \(2y - x^{2} + 4 = 0\) in standard form. Draw a graph of the function and state the significant characteristics.

Write the equation in standard form \(y = ax^{2} + bx + c\)

\begin{align*} 2y - x^{2} + 4 &= 0 \\ 2y &= x^{2} - 4 \\ y &= \frac{1}{2}x^{2} - 2 \end{align*}

Therefore, we see that:

\[a = \frac{1}{2}; \qquad b = 0; \qquad c = -2\]

Draw a graph of the parabola

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

\begin{align*} y &= \frac{1}{2}(0)^{2} - 2 \\ \therefore y &= 0 - 2 \\ &= -2 \end{align*}

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

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

\begin{align*} 0 &= \frac{1}{2}x^{2} - 2 \\ 0 &= x^{2} - 4 \\ 0 &= (x - 2)(x + 2) \\ \therefore x = -2 &\text{ or } x = 2 \end{align*}

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

State the significant characteristics

Shape: \(a > 0\), therefore the graph is a "smile".

Intercepts: \((-2;0), (2;0) \text{ and } (0;-2)\)

Turning point: \((0;-2)\)

Axes of symmetry: \(x = -\frac{b}{2a} = - \frac{0}{2 \left( \frac{1}{2} \right)} = 0\)

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

Range: \(\{ y: y \geq -2, y \in \mathbb{R} \}\)

The function is decreasing for \(x < 0\) and increasing for \(x > 0\).

Worked example 3: Exponential function

Draw the graphs of \(f(x) = 2^{x}\) and \(g(x) = \left( \frac{1}{2} \right)^{x}\) on the same set of axes and compare the two functions.

Examine the functions and determine the information needed to draw the graphs

Consider the function: \(f(x) = 2^{x}\)

\begin{align*} \text{If } y = 0: \quad 2^{x} &= 0 \\ \text{But } 2^{x} &\ne 0 \\ \therefore \text{ no solution} & \\ \text{If } x = 0: \quad 2^{0} &= 1 \\ \text{This gives the point } & (0;1). \end{align*}

Asymptotes: \(f(x) = 2^{x}\) has a horizontal asymptote, the line \(y = 0\), which is the \(x\)-axis.

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
\(f(x)\)	\(\frac{1}{4}\)	\(\frac{1}{2}\)	\(1\)	\(2\)	\(4\)

Consider the function: \(g(x) = \left( \frac{1}{2} \right)^{x}\)

\begin{align*} \text{If } y = 0: \quad \left( \frac{1}{2} \right)^{x} &= 0 \\ \text{But } \left( \frac{1}{2} \right)^{x} &\ne 0 \\ \therefore \text{ no solution}& \\ \text{If } x = 0: \quad \left( \frac{1}{2} \right)^{0} &= 1 \\ \text{This gives the point } & (0;1). \end{align*}

Asymptotes: \(g(x) = \left( \frac{1}{2} \right)^{x}\) also has a horizontal asymptote at \(y = 0\).

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
\(g(x)\)	\(4\)	\(2\)	\(1\)	\(\frac{1}{2}\)	\(\frac{1}{4}\)

Draw the graphs of the exponential functions

State the significant characteristics

Symmetry: \(f\) and \(g\) are symmetrical about the \(y\)-axis.

Domain of \(f\) and \(g\): \(\{ x: x \in \mathbb{R} \}\)

Range of \(f\) and \(g\): \(\{ y: y > 0, y \in \mathbb{R} \}\)

The function \(g\) decreases as \(x\) increases and function \(f\) increases as \(x\) increases. The two graphs intersect at the point \((0;1)\).

Revision

Textbook Exercise 2.1

\(f(x) = 3x^{2}\) and \(g(x) = - x^{2}\)

For \(f(x)\):

\begin{align*} \text{Intercept: } & (0;0) \\ \text{Turning point: } & (0;0) \\ \text{Axes of symmetry: } & x = 0 \\ \text{Domain: } & \{x: x \in \mathbb{R} \} \\ \text{Range: } & \{y: y \geq 0, y \in \mathbb{R} \} \\ \text{Minimum value: } & y = 0 \end{align*}

For \(g(x)\):

\begin{align*} \text{Intercept: } & (0;0) \\ \text{Turning point: } & (0;0) \\ \text{Axes of symmetry: } & x = 0 \\ \text{Domain: } & \{x: x \in \mathbb{R} \} \\ \text{Range: } & \{y: y \leq 0, y \in \mathbb{R} \} \\ \text{Maximum value: } & y = 0 \end{align*}

\(j(x) = - \frac{1}{5}x^{2}\) and \(k(x) = - 5x^{2}\)

For \(j(x)\):

\begin{align*} \text{Intercepts: } & (0;0) \enspace (0;0) \\ \text{Turning point: } & (0;0) \\ \text{Axes of symmetry: } & x = 0 \\ \text{Domain: } & \{x: x \in \mathbb{R} \} \\ \text{Range: } & \{y: y \leq 0, y \in \mathbb{R} \} \\ \text{Maximum value: } & y = 0 \end{align*}

For \(k(x)\):

\(h(x) = 2x^{2} + 4\) and \(l(x) = - 2x^{2} - 4\)

For \(h(x)\):

\begin{align*} \text{Intercept: } & (0;4) \\ \text{Turning point: } & (0;4) \\ \text{Axes of symmetry: } & x = 0 \\ \text{Domain: } & \{x: x \in \mathbb{R} \} \\ \text{Range: } & \{y: y \geq 4, y \in \mathbb{R} \} \\ \text{Minimum value: } & y = 4 \end{align*}

For \(l(x)\):

\begin{align*} \text{Intercept: } & (0;-4) \\ \text{Turning point: } & (0;-4) \\ \text{Axes of symmetry: } & x = 0 \\ \text{Domain: } & \{x: x \in \mathbb{R} \} \\ \text{Range: } & \{y: y \leq -4, y \in \mathbb{R} \} \\ \text{Maximum value: } & y = -4 \end{align*}

Given \(f(x) = -3x - 6\) and \(g(x) = mx +c\). Determine the values of \(m\) and \(c\) if \(g \parallel f\) and \(g\) passes through the point \((1;2)\). Sketch both functions on the same system of axes.

\begin{align*} g(x) &= mx +c \\ m & = -3 \\ g(x) &= -3x +c \\ \text{Substitute } (1;2) \quad 2 &= -3(1) + c \\ \therefore c &= 5 \\ \therefore g(x) &= -3x + 5 \\ \text{Intercepts for } g: \quad & (\frac{5}{3};0);(0;5) \\ & \\ \text{Intercepts for } f: \quad & (-2;0);(0;-6) \end{align*}

Given \(m: \frac{x}{2} - \frac{y}{3} = 1\) and \(n: - \frac{y}{3} = 1\). Determine the \(x\)- and \(y\)-intercepts and sketch both graphs on the same system of axes.

For \(m(x)\):

\begin{align*} \frac{x}{2} - \frac{y}{3} & = 1 \\ \text{Let } x = 0: \quad - \frac{y}{3} & = 1 \\ y & = -3 \\ \text{Let } y = 0: \quad \frac{x}{2} & = 1 \\ x & = 2 \\ \text{Intercepts: } & (2;0);(0;-3) \end{align*}

For \(n(x)\):

\begin{align*} - \frac{y}{3} & = 1 \\ \therefore y & = -3 \\ \text{Intercepts: } & (0;-3) \end{align*}

Write the linear function in standard form:

\begin{align*} \frac{x}{2} - \frac{y}{3} & = 1 \\ \frac{3x}{2} - y & = 3 \\ \frac{3x}{2} - 3 & = y \\ \therefore y &= \frac{3x}{2} - 3 \end{align*}

Sketch \(p, q \text{ and } r\) on the same system of axes.

For each of the functions, determine the intercepts, asymptotes, domain and range.

For \(p(x)\):

\begin{align*} \text{Intercept: } & (0;1) \\ \text{Asymptote: } & y = 0 \\ \text{Domain: } & \{x: x \in \mathbb{R} \} \\ \text{Range: } & \{y: y > 0 \} \end{align*}

For \(q(x)\):

\begin{align*} \text{Intercept: } & (0;1) \\ \text{Asymptote: } & y = 0 \\ \text{Domain: } & \{x: x \in \mathbb{R} \} \\ \text{Range: } & \{y: y > 0 \} \end{align*}

For \(r(x)\):

\begin{align*} \text{Intercept: } & (0;-1) \\ \text{Asymptote: } & y = 0 \\ \text{Domain: } & \{x: x \in \mathbb{R} \} \\ \text{Range: } & \{y: y < 0 \} \end{align*}