# 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

#### Questions & Answers

The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
divide simplify each answer 3/2÷5/4
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
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
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
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