# 2.2 Graphs of linear functions  (Page 4/15)

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Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point $\left(-2,0\right).$ To get from this point to the y- intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

$m=\frac{\text{rise}}{\text{run}}=\frac{4}{2}=2$

Substituting the slope and y- intercept into the slope-intercept form of a line gives

$y=2x+4$

Given a graph of linear function, find the equation to describe the function.

1. Identify the y- intercept of an equation.
2. Choose two points to determine the slope.
3. Substitute the y- intercept and slope into the slope-intercept form of a line.

## Matching linear functions to their graphs

Match each equation of the linear functions with one of the lines in [link] .

1. $f\left(x\right)=2x+3$
2. $g\left(x\right)=2x-3$
3. $h\left(x\right)=-2x+3$
4. $j\left(x\right)=\frac{1}{2}x+3$

Analyze the information for each function.

1. This function has a slope of 2 and a y -intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function $g$ has the same slope, but a different y- intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through so $f$ must be represented by line I.
2. This function also has a slope of 2, but a y -intercept of $-3.$ It must pass through the point $\left(0,-3\right)$ and slant upward from left to right. It must be represented by line III.
3. This function has a slope of –2 and a y- intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.
4. This function has a slope of $\frac{1}{2}$ and a y- intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through but the slope of $j$ is less than the slope of $f$ so the line for $j$ must be flatter. This function is represented by Line II.

Now we can re-label the lines as in [link] .

## Finding the x -intercept of a line

So far, we have been finding the y- intercepts of a function: the point at which the graph of the function crosses the y -axis. A function may also have an x -intercept, which is the x -coordinate of the point where the graph of the function crosses the x -axis. In other words, it is the input value when the output value is zero.

To find the x -intercept, set a function $f\left(x\right)$ equal to zero and solve for the value of $x.$ For example, consider the function shown.

$f\left(x\right)=3x-6$

Set the function equal to 0 and solve for $x.$

$\begin{array}{l}0=3x-6\hfill \\ 6=3x\hfill \\ 2=x\hfill \\ x=2\hfill \end{array}$

The graph of the function crosses the x -axis at the point

Do all linear functions have x -intercepts?

No. However, linear functions of the form $y=c,$ where $c$ is a nonzero real number are the only examples of linear functions with no x-intercept. For example, $y=5$ is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in [link] .

## x -intercept

The x -intercept of the function is value of $x$ when $f\left(x\right)=0.$ It can be solved by the equation $0=mx+b.$

## Finding an x -intercept

Find the x -intercept of $f\left(x\right)=\frac{1}{2}x-3.$

Set the function equal to zero to solve for $x.$

$\begin{array}{l}0=\frac{1}{2}x-3\\ 3=\frac{1}{2}x\\ 6=x\\ x=6\end{array}$

The graph crosses the x -axis at the point

how can are find the domain and range of a relations
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
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more than 6000
Robert
can I see the picture
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with doing calculus
SLIMANE
Thanks po.
Jenica
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Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
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SLIMANE
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This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
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Jeffrey
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meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
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Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
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