Algebraic Representation of a Function - Examples, Exercises and Solutions

It should be remembered that a constant function describes a situation where as the X value increases, the function value (Y) remains constant.

In the table, we can see that there is a constant change in X values, meaning an increase of 1, and a non-constant change in Y values - sometimes increasing by 1 and sometimes by 4

Therefore, according to the rule, the table does not describe a function

It is important to remember that a constant function describes a situation where as the X value increases, the function value (Y) remains constant.

In the table, we can see that there is a constant change in X values, meaning an increase of 1, and a constant change in Y values, meaning an increase of 3

Therefore, according to the rule, the table describes a function.

It is important to remember that a constant function describes a situation where as the X value increases, the function value (Y) remains constant.

In the table, we can see that there is a constant change in the X values, specifically an increase of 2, and the Y value remains constant.

Therefore, according to the rule, the table describes a constant function.

It is important to remember that a function is an equation that assigns to each value in domain $x$ only one value in range $y$.

Since we can see that for every $x$ value found on the graph there is only one corresponding$y$ value, the graph is indeed a function.

It is important to remember that a function is an equation that assigns to each element in domain X one and only one element in range Y

We should note that for every X value found on the graph, there is one and only one corresponding Y value.

Therefore, the graph is indeed a function.

It is important to remember that a function is an equation that assigns to each element in domain X one and only one element in range Y

Let's note that in the graph:

$f(0)=2,f(0)=-2$

In other words, there are two values for the same number.

Therefore, the graph is not a function.

Let's use the below formula in order to find the slope:

$m=\frac{y_2-y_1}{x_2-x_1}$

We begin by inserting the known data from the graph into the formula:

$(0,-2),(-2,0)$

$m=\frac{-2-0}{0-(-2)}=$

$\frac{-2}{0+2}=$

$\frac{-2}{2}=-1$

We then substitute the point and slope into the line equation:

$y=mx+b$

$0=-1\times(-2)+b$

$0=2+b$

Lastly we combine the like terms:

$0+(-2)=b$

$-2=b$

Therefore, the equation will be:

$y=-x-2$

We will begin by using the formula for finding slope:

$m=\frac{y_2-y_1}{x_2-x_1}$

First let's take the points:

$(-1,4),(3,8)$

$m=\frac{8-4}{3-(-1)}=$

$\frac{8-4}{3+1}=$

$\frac{4}{4}=1$

Next we'll substitute the point and slope into the line equation:

$y=mx+b$

$8=1\times3+b$

$8=3+b$

Lastly we'll combine like terms:

$8-3=b$

$5=b$

Therefore, the equation will be:

$y=x+5$