From a straight line to the function equation

A common problem is the following one. We are looking for the equation of a linear function $f$ , but all we know are the coordinates of two points through which the graph of $f$ passes. Such a graph is shown below. It passes through the points $A(1\vert 3)$ and $B(6\vert 7)$ . How do we find its function equation

f(x)=ax+b

Here is how it works. We can use the two points to find the slope $a$ by choosing the slope triangle as shown in the figure above. $\Delta x$ then tells you how to get from the $x$ -coordinate of $A$ to the $x$ -coordinate of $B$

\Delta x = 6-1=5

and $\Delta y$ tells you how to get from the $y$ -coordinate of $A$ to the $y$ -coordinate of $B$

\Delta y = 7-3=4

Thus we have

a=\frac{\Delta y}{\Delta x}=\frac{4}{5}=0.8

We therefore have

f(x)=0.8x+b

How can we find $b$ ? Well, we know that the graph passes through $A$ , so we know that

f(1)=3 \rightarrow 0.8\cdot 1+b=3 \rightarrow b=2.2

Of course we could have taken $B$ as well, and get the same $b$ :

f(6)=7 \rightarrow 0.8\cdot 6+b=7 \rightarrow b=2.2

The function equation of $f$ is therefore

f(x)=0.8x+2.2

Exercise 1

Q1

Find the function equation of a linear function $f$ whose graph

passes through the points $U(1|4)$ and $V(5|6)$
passes through the points $U(-3|3.2)$ and $V(2|-10)$
has an $y$ -intercept at $-3.3$ and crosses the $x$ -axis at $-4.4$
intersects the $y$ -axis at $4$ and the $x$ -axis at $7$
is the diagonal of a square of side length $2$ (two solutions), where the lower left point of the square is at the origin.
has $y$ -intercept $-3$ and slope $-2$ .

Q2

Determine the point of intersection $S$ between the straight lines $g$ and $h$

Q3

Consider the linear function $f(x)=0.5x+1$ . Find another linear function whose graph forms a right angle with the graph of $f$ and passes through the point $P(4|3)$ .

Solution

A1

$f(x)=ax+b$ . Draw a figure!!

$a=\frac{\Delta y}{\Delta x}=\frac{2}{4}=0.5$ , thus $f(x)=0.5x+b$ . $f(1)=4 \rightarrow 0.5\cdot 1+b=4 \rightarrow b=3.5$ . Thus $f(x)=\underline{0.5x+3.5}$ .
$a=\frac{\Delta y}{\Delta x}=\frac{-13.2}{5}=-2.64$ , thus $f(x)=-2.64x+b$ . $f(2)=-10 \rightarrow 2\cdot (-2.64)+b=-10 \rightarrow b=-4.72$ . Thus $f(x)=\underline{-2.64x-4.72}$ .
$U(-4.4|0)$ , $V(0|-3.3)$ . $a=\frac{\Delta y}{\Delta x}=\frac{-3.3}{4.4}=\underline{-0.75}$ , thus $f(x)=-0.75x+b$ . $b=-3.3$ ( $y$ -intercept). Thus $f(x)=\underline{-0.75x-3.3}$ .
$U(0|4)$ , $V(7|0)$ $a=\frac{\Delta y}{\Delta x}=\frac{-4}{7}$ , thus $f(x)=-\frac{4}{7}x+b$ . $f(0)=4=b$ , thus $f(x)=\underline{-\frac{4}{7}x+4}$ .
The square is shown below.Solution 1: graph passes through $U(0\vert 2)$ and $V(2\vert 0)$ , thus $f(x)=\underline{-x+2}$ . Solution 2: graph passes through $U(0\vert 0)$ and $V(2\vert 2)$ , thus $f(x)=\underline{x}$ .
$f(x)=\underline{-2x-3}$ .

A2

First, find the function equations of $h$ and $g$ . $g(x)=ax+b$ with slope $a=\frac{4}{3}$ , so $g(x)=\frac{4}{3}x+b$ . As point $P(1|1)$ is on $g$ , it follows $g(1)=1\rightarrow \frac{4}{3}\cdot 1+b=1 \rightarrow b=-\frac{1}{3} \rightarrow g(x)=\frac{4}{3}x-\frac{1}{3}$ . $h(x)=ax+b$ with slope $a=-\frac{3}{3}=-1$ , so $h(x)=-x+b$ . As point $P(1|5)$ is on $h$ , it follows $h(1)=5\rightarrow -1+b=5 \rightarrow b=6 \rightarrow h(x)=-x+6$ . Second, we can find the $x$ -coordinate of $S$ : Find $x$ such that

g(x)=h(x)

\frac{4}{3}x-\frac{1}{3}=-x+6

\frac{7}{3}x=\frac{19}{3} \rightarrow 7x=19 \rightarrow x=\frac{19}{7}=2.71... .

The $y$ -coordinate is given by $y=h(\frac{19}{7})=-\frac{19}{7}+6=\frac{23}{7}=3.28...$ . We could have taken the function $g$ as well, but this will get the same $y$ -coordinate. The point $S$ has therefore the coordinates $\underline{S(\frac{19}{7}|\frac{23}{7})}$ .

A3

Let us call the new function $g$ , and the function equation is $g(x)=ax+b$ . From the figure (see below) it follows that the graph of $g$ has slope $a=\frac{-4}{2}=-2$ and because it passes through the point $P(4|3)$ , it is $g(4)=3\rightarrow -2\cdot 4+q=3\rightarrow q=11 \rightarrow \underline{g(x)=-2x+11}$ .