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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

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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
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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?
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lim x to infinity e^1-e^-1/log(1+x)
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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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