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y = 2 x + 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] .

a .   f ( x ) = 2 x + 3 b . g ( x ) = 2 x 3 c . h ( x ) = −2 x + 3 d j ( x ) = 1 2 x + 3
Graph of four functions where the orange line has a y-intercept at 3 and slope of 2, the baby blue line has a y-intercept at 3 and slope of 1/2, the blue line has a y-intercept at 3 and slope of -2, and the green line has a y-intercept at -3 and slope of 2.

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 ( 0 , 3 ) 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 ( 0 , −3 ) 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 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 ( 0 , 3 ) , 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] .

Graph of four functions where the blue line is h(x) = -2x + 3 which goes through the point (0,3), the baby blue line is j(x) = x/2 + 3 which goes through the point (0,3). The orange line is f(x) = 2x + 3 which goes through the point (0,3), and the red line is g(x) = 2x – 3 which goes through the point (0,-3).
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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. Recall that 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 ( x ) equal to zero and solve for the value of x . For example, consider the function shown.

f ( x ) = 3 x 6

Set the function equal to 0 and solve for x .

0 = 3 x 6 6 = 3 x 2 = x x = 2

The graph of the function crosses the x -axis at the point ( 2 , 0 ) .

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] .

Graph of the function y = 5, a completely horizontal line that goes through the point (0,5).  Graphed on an xy-plane with the x-axis ranging from -3 to 3 and the y-plane ranging from -1 to 8.

x -intercept

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

Finding an x -intercept

Find the x -intercept of f ( x ) = 1 2 x 3.

Set the function equal to zero to solve for x .

0 = 1 2 x 3 3 = 1 2 x 6 = x x = 6

The graph crosses the x -axis at the point ( 6 , 0 ) .

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Find the x -intercept of f ( x ) = 1 4 x 4.

( 16 ,  0 )

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Describing horizontal and vertical lines

There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line    indicates a constant output, or y -value. In [link] , we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 in the equation f ( x ) = m x + b , the equation simplifies to f ( x ) = b . In other words, the value of the function is a constant. This graph represents the function f ( x ) = 2.

Questions & Answers

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Adu
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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