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# Interpreting Linear Functions

Linear functions are those functions which forms a straight line when drawn on a graph sheet. The most common form of linear functions is $y$ = $mx + b$, where $x$ and $y$ are variables, $m$ is called the slope or gradient and $b$ is the $y$ intercept. Interpreting linear functions would understand the information given in a problem and framing the linear equation based on the data provided, also drawing appropriate graph of the linear function of it.

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## Definition

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Linear functions can be defined as a function having a combination of two variables and constants. The values of the exponent that any of the variables could carry are either $0$ or $1$. There are primarily three forms of a linear function which are slope intercept form, point slope form and the general form. The slope intercept form is the most common form of representing any linear function. It is $y$ = $mx + b$ where $m$ is the measure of steepness called the slope or the gradient and $b$ is the point at which the line cuts the $y$ axis whereby the value of $x$ = $0$ and is called the $y$ intercept. The next common form is the point slope form. It is $(y - y1)$ = $m \times (x - x1)$, where $(x, y)$ and $(x1, y1)$ are the coordinates of any two points lying on the linear function and $m$ is the slope of the line which is rise over run. The third form is the general form $Ax + By + C$ = $0$ where $A$ and $B$ are the coefficients such that they cannot be equal to $0$ and $C$ is a constant.

Interpreting linear function can be defined as explaining the characteristics of a linear function given and drawing the related graph on it.

## Constructing and Interpreting Linear Functions

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When a situation is given and we are asked to form a linear function on it and also sketch a graph related to the situation, then it is called constructing and interpreting linear functions. The best way to explain it would be with the help of an example as follows:

Example:

In a race the tyres of a car gets worn out. It loses $3 mm$ of tread on completing $250 km$ of race and $4 mm$ of race on completing $1000 km$ of race. It is known that the loss of tread is directly proportional to the distance covered in the race. Based on the information given, find out the following parts:

i) On covering a distance of $s\ km$ what would be the loss of tread $d\ mm$?

ii) On covering a distance of $2000 km$ race, what shall be the amount of loss?

iii) Sketch a graph of the linear function formed in part a

Solution:

i) On completing a distance of $250 km$, loss of tread is $3mm$ and on completing a distance of $1000 km$, loss of tread is $4 mm$. So, the gradient or slope would be,

$'m$ = $\frac{Rise}{Run}$

$'m$ = $\frac{4 - 3}{1000 - 250}$

$'m$ = $\frac{1}{750}$

After finding the slope we substitute that value in the standard form of linear function

$'d$ = $ms + c$

Where, $s$ is the distance covered, $m$ is the slope = $\frac{1}{750}$, $d$ is the loss of tread and $c$ is the constant or $y$ intercept. Plugin the first pair of coordinate given $s$ = $250 km$ and $d$ = $3 mm$, we get

$3$ = $\frac{1}{750}$ $\times 250 + c$

$3$ = $\frac{1}{3 + c}$

$\frac{3 - 1}{3}$ = $c$

$'c$ = $2$ $\frac{2}{3}$

Thus, the linear function formed is $d$ = $\frac{s}{750}$ + $\frac{8}{3}$. Simplifying it further,

$'d$ = $\frac{(s + 2000)}{750}$

$750d$ = $s + 2000$

ii) We plugin the value of $s$ = $2000 km$ in the linear equation in part i,

$750 d$ = $2000 + 2000$

$750 d$ = $4000$

$'d$ = $\frac{4000}{750}$

$'d$ = $\frac{51}{3}$

Thus, the loss of trade on covering a distance of $2000 km$ is $\frac{51}{3}$ mm.

iii) ## Interpreting a Table of Linear Functions

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When we try to draw the graph of a linear function we need the coordinates to plot. Those coordinates are found out by assuming few values of $x$. Then, we substitute those values in the given linear function and solve for the corresponding $y$ values of $y$ coordinate. This is done in a table format and hence is known as interpreting a table of linear functions.

## Interpreting Linear Functions Graphs

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When a graph is given and we are asked to interpret, it means to understand the relation between the property which represents the $y$ coordinate and the property that represents the $x$ coordinate. If the graph is a straight line, it mean it is a linear function. If the straight line is inclined in an upward direction it shows they have positive slope, with the increase of $x$ coordinate the $y$ coordinate also increases. If the graph or the straight line is inclined in a downward position it means they have negative slope, with the increase of $x$ coordinates the $y$ coordinates are decreasing.

## Word Problems

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Example 1:

Lucy would like to enroll herself in a Spanish class. The one time enrollment fee is $\$100$and thereafter every week the fees are$ \$5$. Find out the linear function between the enrollment fee and the number of weeks she enrolled for. Also, find out her total cost of enrollment after attending $8$ weeks of Spanish class

Solution:

Let the number of weeks Lucy has enrolled for her Spanish class be $x$. If per week enrollment fee is $\$5$, then the enrollment fee for$x$weeks shall be$ \$5x$. Let the total cost of enrollment be represented by $y$. So, $y$ = $100 + 5x$ is the linear function between the enrollment fee and the number of weeks she enrolled for.

Substituting the value of $x$ = $8$ in the linear function, we get

$y$ = $100 + 5x$

$y$ = $100 + 5 \times 8$

$y$ = $100 + 40$

$y$ = $\$140$Thus, the total cost of enrollment after attending$8$weeks of Spanish class is$ \$140$.

Example 2:

Below is given a table containing $x$ and $f(x)$ values such that they form a linear function. Few of the values are missing. Complete the table by finding out the missing numbers.

X -3 0 - 4 7 -
F(x) 17 - 1 -18 - -30

Solution:

As the relation between $x$ and $f(x)$ forms a linear function, it obeys the slope intercept form $y$ = $ax + b$. Minimum two pairs of coordinates are required to form the equation as number of variables is two in number. Looking at the table we got two ordered pairs of coordinates as $(-3, 17)$ and $(4, -18)$.

Let us substitute the values of $x$ and $y$ in $y$ = $ax + b$,

$17$ = $a \times -3 + b$

$17$ = $-3a + b$ ....... equation 1

$-18$ = $a \times 4 + b$

$-18$ = $4a + b$ ......... equation 2

Using the method of elimination to solve the equations, we subtract equation $1$ from equation $2$

$-18$ = $4a + b$

$(-) 17$ = $(-) -3a (-) + b$

$-35$ = $7a$

$'a$ = $-5$

Plugin the value of $a$ = $-5$ in equation $1$, we get

$17$ = $-3a + b$

$17$ = $-3 \times -5 + b$

$17$ = $15 + b$

$B$ = $2$

Substituting the values of $a$ and $b$ in the general form $y$ = $ax + b$, we get $y$ = $-5x + 2$. There are four missing values, two of $x$ and two of $f(x)$.

When $x$ = $0$, $y$ = $-5 \times 0 + 2\ \Rightarrow\ y$ = $2$

When $f(x)$ = $1,\ 1$ = $-5x + 2 \Rightarrow\ -1$ = $-5x \Rightarrow\ x$ = $\frac{1}{5}$ $\Rightarrow\ x$ = $0.2$

When $x$ = $7,\ y$ = $-5 \times 7 + 2 \Rightarrow\ y$ = $-33$

When $f(x)$ = $-30,\ -30$ = $-5x + 2 \Rightarrow\ x$ = $\frac{-32}{-5}$ $\Rightarrow\ x$ = $6.4$

Thus, the complete table shall look like this:

X -3 0 0.2 4 7 6,4
F(x) 17 2 1 -18 -33 -30
Example 2:

Lucy has planted an aloe Vera plant and examines its growth. When planted its length was $5 cm$ and every fortnight it grows by $1.5 cm$. Assuming that the growth of the plant forms a linear function, find out the following:

a) Form the linear function relating the height of the plant $(h)$ after '$d$' number of days have passed out

b) Find out the length of the plant after $20$ days have passed out

Solution:

a) The slope intercept form of linear function is $y$ = $mx + b$. Here $m$ or slope is the rate at which the plant is growing = $1.5 cm$ and the $y$ intercept is $b$ = $5 cm$. Also $x$ = '$d$' number of days and $y$ = '$h$' height of the plant. Substituting these values in the slope intercept form, we get

$'h$ = $1.5d + 5$

Thus, the linear function relating the height of the plant $(h)$ after '$d$' number of days have passed out is $h$ = $1.5d + 5$

b) Substituting the value of $d$ = $20$ in the linear function formed in part a) $h$ = $1.5d + 5$, we get $h$ = $1.5 \times 20 + 5 \Rightarrow\ h$ = $35$. So, the length of the plant after $20$ days have passed out is $35$ cms.
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