Further exercises: Linear functions

1
The straight line $y = - 3 x$ is mirrored on the $x$ -axis.
The mirrored straight line is then shifted upwards by 2 units.
Determine the equation of the new straight line;
a) by a drawing
b) by computation

Draw the function $y = - 3 x$
Mirror the line along the $x$ -axis.
Read the new equation of the function.
It is: $y = 3 x$
Move the straight line up by 2 units.
The $y$ -axis intercept has changed.
Read off the new function equation.
Answer: The straight line mirrored on the $x$ -axis and shifted upwards by $2$ units has the equation $y = 3 x + 2$
Solution for b)
When mirroring on the $x$ -axis, the function term is multiplied by $(- 1)$ .
$y = (- 1) \cdot (- 3 x) = 3 x$
Thus the equation of the mirrored straight line is $y = 3 x $
Move this straight line up by 3.
This corresponds to the $y$ -axis intercept $t \Rightarrow$ $t = 2$
Answer: The new line equation is:
$y = 3 x + 2$
Solution for a)
2
Consider the functions $g (x) = 0.75 x + 3$ and $h (x) = - x - 2.5$ .
The straight line $h$ is to be shifted in the $y$ -direction so that $g$ and the shifted line $h$ intersect the $x$ -axis at the same point.
Determine the function term $f (x)$ for the shifted line.

For this task you need the following basic knowledge: Linear function
As you can see from the picture, you have to move the line $h$ in the $y$ -direction so that $f$ and $g$ then have the same zero.
$f$ has the same slope as $h$ , i.e. $m = - 1$ .
Determine the zero of $g$
$0$ $=$ $\frac{3}{4} \cdot x + 3$
↓
solve for x
$x$ $=$ $- 4$
Determine the line equation of $f$
Insert the zero point $(- 4 | 0)$ and the slope of $f$ into the general line equation.
$y$ $=$ $m \cdot x + t$
$0$ $=$ $(- 1) \cdot (- 4) + t$
$t$ $=$ $- 4$
So the line equation for $f$ is:
$f (x) = - x - 4$
3
Consider the point $P (t | \frac{t}{2} + 3)$ with $t \in ℝ$
Choose some values for $t$ and plot the corresponding points in a coordinate system.
How are the points located in the coordinate system? For which $t$ - values is the $x$ -coordinate equal to the $y$ -coordinate of point $P$ ?

For example, set $t = {0; 1; 2; 3; 4}$ and calculate the corresponding $y$ -values. Other numbers for $t$ are also admitted.
$t = 0 : y$ $=$ $\frac{0}{2} + 3$
$y$ $=$ $3$
So we have the point $A (0 | 3)$ .
$t = 1 : y$ $=$ $\frac{1}{2} + 3$
$y$ $=$ $3.5$
So we have the point $B (1 | 3.5)$ .
$t = 2 : y$ $=$ $\frac{2}{2} + 3$
$y$ $=$ $4$
So we have the point $C (2 | 4)$ .
$t = 3 : y$ $=$ $\frac{3}{2} + 3$
$y$ $=$ $4.5$
So we have the point $D (3 | 4.5)$ .
$t = 4 : y$ $=$ $\frac{4}{2} + 3$
$y$ $=$ $5$
So we have the point $E (4 | 5)$ .
Draw the points A, B, C, D and E in a coordinate system.
$\Rightarrow$ All points lie on a common line. This line has the equation $y = \frac{t}{2} + 3$
Calculate when the $x$ and $y$ coordinates have the same value.
Set $y$ equal to $t$ for this.
$t = y$
Replace $y$ by the term $\frac{t}{2} + 3$ .
$t$ $=$ $\frac{t}{2} + 3$ $\cdot 2$
$2 t$ $=$ $t + 6$ $- t$
$t$ $=$ $6$
$\Rightarrow$ The $x$ - and $y$ - coordinates have the same value when $t = 6$ . The corresponding point is $P (6 | 6)$ .
4
Show that the line $g$ through $P_{1} (\sqrt{k} / k)$ and $P_{2} (1 / 1)$ has the slope $a_{1} = \sqrt{k} + 1$ and intersects the y-axis at $P_{y} (0 / - \sqrt{k})$

For this task you need the following basic knowledge: Line equation
Plug $P_{1}$ and $P_{2}$ into the general line equation.
$y = m \cdot x + t$
1) $k = m \cdot \sqrt{k} + t$
2) $1 = m \cdot 1 + t$
$1) - 2) :$
Use the addition method.
$k - 1$ $=$ $ m \sqrt{k} - m + t - t$
↓
conclude
$k - 1$ $=$ $m (\sqrt{k} - 1)$
↓
solve for $m$
$m$ $=$ $\frac{k - 1}{\sqrt{k} - 1}$
↓
simplify using the binomial theorems
( $k = {\sqrt{k}}^{2})$ .
$m$ $=$ $\frac{(\sqrt{k} + 1) (\sqrt{k} - 1)}{\sqrt{k} - 1}$
$m$ $=$ $\sqrt{k} + 1$
↓
use 2)
$1$ $=$ $(\sqrt{k} + 1) \cdot 1 + t$
↓
solve for $t$
$t$ $=$ $1 - (\sqrt{k} + 1)$
$t$ $=$ $- \sqrt{k}$
$\Rightarrow$ The slope really is $a_{1} = \sqrt{k} + 1$ and the line intersects the $y$ -axis at $P_{y} (0 | - \sqrt{k}) .$
5
Show that the points $P (\frac{k}{2} \sqrt{2} / k)$ lie on a straight line for all $k \in ℝ$ .
Determine the equation of the straight line.

For this task you need the following basic knowledge: Linear function
Consider the set of points $P (\frac{k}{2} \cdot \sqrt{2} | k)$ for different $k$ .
To find the general equation of a straight line, we need two points on the straight line.
Therefore, it is best to choose here e.g. the two points $F (\frac{k}{2} \cdot \sqrt{2} | k)$ and $G (\frac{k + 1}{2} \cdot \sqrt{2} | k + 1)$ .
y $= m \cdot x + t$
First determine the slope $m$ .
$m = \frac{k + 1 - k}{\frac{k + 1}{2} \cdot \sqrt{2} - \frac{k}{2} \cdot \sqrt{2}} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}}$
Now you can insert $m$ and one of the points, e.g. $P$ , into the general line equation.
$k = \underset{= k}{\underset{⏟}{\frac{k}{2} \cdot \sqrt{2} \cdot \frac{2}{\sqrt{2}}}} + t$
Solve for $t$ .
$\Rightarrow$ t=0
Wrtite down the equation of the straight line.
$y = \frac{2}{\sqrt{2}} \cdot x$
6
Find the function term of the linear function $f (x)$ for:
$f (1) = 7; f (- 1) = 3$

For this task you need the following basic knowledge: Linear function
Substitute $f (1)$ and $f (- 1)$ into the general equation of a straight line.
$\begin{array}{l} 1) 7 = m \cdot 1 + t \\ 2) 3 = m \cdot (- 1) + t \end{array}$
Solve the system of equations.
$1) + 2) : 10 = 2 t | : 2$
$\Rightarrow t = 5$
Plug $t$ into $1)$
$7 = m + 5 | - 5$
$m = 2$
Substitute $m$ and $t$ into the general equation of a straight line.
$f (x) = 2 x + 5$
$f (x) = m \cdot x + t$

$f (a) = 0; f (0) = a$

For this task you need the following basic knowledge: Linear function
$f (x) = m \cdot x + t$
$t$ is the $y$ -axis intercept, i.e. the value at $x = 0$ .
$t = f (0) = a$
Substitute this and $f (a) = 0$ into the general line equation.
$0 = m \cdot a + a$
Solve for $m$ .
$m = \frac{- a}{a} = - 1$
Plug $m = - 1$ and $t = a$ into the general equation of the function.
$f (x) = - x + a$

$f (a) = 1; f (2 a) = - 1$

For this task you need the following basic knowledge: Linear function
$f (x) = m \cdot x + t$
Substitute $f (a)$ and $f (2 a)$ into the general equation of a straight line.
$\begin{array}{rrclcr} I & 1 & = & m \cdot a & + & t \\ II & - 1 & = & m \cdot 2 a & + & t \end{array}$
Solve the system of equations.
$\begin{array}{l} 2 \cdot I - II : \\ 2 - (- 1) = 2 t - t \end{array}$
$\Rightarrow t = 3$
Plug $t$ into $I$ .
$1 = m \cdot a + 3$
Solve the equation for $m$ .
$m = (1 - 3) \cdot \frac{1}{a} = - \frac{2}{a}$
Substitute $m = - \frac{2}{a}$ and $t = 3$ into the general line equation.
$f (x) = - \frac{2}{a} \cdot x + 3$
7
Determining intersection points
Three lines are drawn in the coordinate system. Read off the points of intersection from the figure.

For this task you need the following basic knowledge: System of linear equations
Reading off intersection points
In the picture on the right you can determine the approximate coordinates of the three points. They are:
$S_{1} (- 2 | 4)$
$S_{2} (2 | 3)$
$S_{3} (- 1.5 | - 2)$
8
Determining intersection points
Consider a line $g$ and a line $h$ .
Determine the line equations of $g$ and $h$ .

For this task you need the following basic knowledge: Line equation
$y = m x + t$
The parameter $m$ is the slope of the straight line. The parameter $t$ is the $y$ -axis intercept.
Line g:
$y = m_{g} x + t_{g}$
The line $g$ has the gradient $- \frac{3}{2}$ , which means $m_{g} = - \frac{3}{2}$ .
$y = - \frac{3}{2} x + t_{g}$
It intersects the $y$ -axis at $\frac{9}{2}$ , which means $t_{g} = \frac{9}{2}$ .
$y = - \frac{3}{2} x + \frac{9}{2}$
Line h:
$y = m_{h} x + t_{h}$
The line $h$ has the gradient $2$ , which means $m_{h} = 2$
$y = 2 x + t_{h}$
It intersects the $y$ -axis at $1$ , which means $t_{h} = 1$ .
$y = 2 x + 1$

Lies den Schnittpunkt ab.

For this task you need the following basic knowledge: Lines in coordinate systems
Reading off intersection points
From the drawing we read off the intersection point $SP (1 | 3)$

$y$	$=$	$m \cdot x + t$
$0$	$=$	$(- 1) \cdot (- 4) + t$
$t$	$=$	$- 4$

$t$	$=$	$\frac{t}{2} + 3$	$\cdot 2$
$2 t$	$=$	$t + 6$	$- t$
$t$	$=$	$6$

$k - 1$	$=$	$ m \sqrt{k} - m + t - t$
	↓	conclude
$k - 1$	$=$	$m (\sqrt{k} - 1)$
	↓	solve for $m$
$m$	$=$	$\frac{k - 1}{\sqrt{k} - 1}$
	↓	simplify using the binomial theorems ( $k = {\sqrt{k}}^{2})$ .
$m$	$=$	$\frac{(\sqrt{k} + 1) (\sqrt{k} - 1)}{\sqrt{k} - 1}$
$m$	$=$	$\sqrt{k} + 1$
	↓	use 2)
$1$	$=$	$(\sqrt{k} + 1) \cdot 1 + t$
	↓	solve for $t$
$t$	$=$	$1 - (\sqrt{k} + 1)$
$t$	$=$	$- \sqrt{k}$

Further exercises: Linear functions

Solution for b)

Solution for a)

Determine the zero of g

Determine the line equation of f

Determining intersection points

Reading off intersection points

Determining intersection points

Reading off intersection points

Determine the zero of $g$

Determine the line equation of $f$