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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 ( 2 , 0 ) . 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 = rise run = 4 2 = 2

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

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] .

  1. f ( x ) = 2 x + 3
  2. g ( x ) = 2 x 3
  3. h ( x ) = 2 x + 3
  4. j ( x ) = 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 ( 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] .

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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. 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 y = 5.

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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Questions & Answers

how to understand calculus?
Jenica Reply
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.
rachel Reply
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.
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
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).
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?
Reena Reply
what is foci?
Reena Reply
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.
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
Bryssen Reply
i want to sure my answer of the exercise
meena Reply
what is the diameter of(x-2)²+(y-3)²=25
Den Reply
how to solve the Identity ?
Barcenas Reply
what type of identity
Confunction Identity
how to solve the sums
hello guys
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.
Shakeena Reply
by how many trees did forest "A" have a greater number?
how solve standard form of polar
Rhudy Reply
what is a complex number used for?
Drew Reply
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
I would like to add that they are used in AC signal analysis for one thing
Good call Scott. Also radar signals I believe.
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
Is there any rule we can use to get the nth term ?
Anwar Reply
how do you get the (1.4427)^t in the carp problem?
Gabrielle Reply
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
ayesha Reply
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Sandra Reply
Practice Key Terms 5

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