Science:Math Exam Resources/Courses/MATH110/December 2010/Question 06 (a)

MATH110 December 2010

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Question 06 (a)
A circle of radius 5 centered on the origin has the equation $\displaystyle x^{2}+y^{2}=25$ This circle is not a function. Why not?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
For a relation to be a function, how many y-values can be associated with a single x-value?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Given a relation or rule that takes input values (usually denoted x) and produces output values (usually denoted y), the relation is a function if each x-value is associated with only one y-value. In other words, the graph of the relation must pass the "vertical line test" - if you can draw a vertical line that intersects the graph at two points, the relation is not a function. In this case, a single x-value in the domain will be associated with two y-values from the upper and lower half of the circle. This is the same as saying that a circle fails the vertical line test, as a vertical line drawn through any point in the domain will intersect the graph in two places.

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