At first glance, the given function looks very intimidating due to its length and the inclusion of many fractions. One should realize, however, that if like terms are combined, the equation quickly condenses to the standard form of a line. Additionally, the concept of an "x-intercept" might not be immediately familiar, but the student should intuit that the x-intercept is the value of x, where the line crosses the x-axis. Another way of saying this is, "the x-intercept is the x value when y=0". Therefore, plug zero in for "y" and eliminate those "y" terms immediately. That leaves: 

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The quickest way to finish this problem is to convert all fractions on the left-hand side to decimal form. (A student should quickly recognize that all fractions on the left-hand side are easily converted to decimals even without the use of a calculator).

The conversion to decimal form yields: 

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Now, combine the like x terms to obtain: 

.

Finally, divide both sides of this equation by -2, to get:

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Recall that this is the value of x when y=0 (because we have already plugged zero in for y) and therefore, this answer represents the x-intercept of the original line.

Which of the following points is on the line of f(x)?

We can solve this problem by plugging in all of the options and seeing which one works. However, we can probably work more quickly by trying the easier options first.

Let's begin with the options including 0, 0 usually makes an equation easier to look at by simplifying things.

So  is out.

Next up, try 

We are good, the point  is on our line, so it is our answer.

To determine if a point works, plug it in and see if it makes a true statement.

The correct answer does:

Answers that don't work include :

NOT TRUE.

In order to determine which point will satisfy the equation, we will have to plug in each value of , solve the right side of the equation, and see if the  will match for the order pairs given.

The only order pair that will satisfy this criteria is  since:

The answer is:  

The -coordinate of the -intercept of a line can be found by substituting 0 for  in its equation and solving for :

The -intercept is the point .

The -coordinate of the -intercept of a line can be found by substituting 0 for  in its equation and solving for :

The -intercept is the point .

The -coordinate of the -intercept of a line can be found by substituting 0 for  in its equation and solving for :

The -intercept is the point .

The -coordinate of the -intercept of a line can be found by substituting 0 for  in its equation and solving for :

The -intercept is the point .

The -coordinate of the -intercept of a line can be found by substituting 0 for  in its equation and solving for :

The -intercept is the point .

The -coordinate of the -intercept of a line can be found by substituting 0 for  in its equation and solving for :

.

The -intercept is the point .