Interpretation of Graph - Fcns of Polynomials: triangle area as sqr root of poly fcn

[phob16]

Hello! I'm struggling to work out how to interpret this graph as attached with equations and questions.

I've managed to conclude that the zeros create a restriction for the domain that the side length C can exist as. The magnitude of side length C is represented by the x-axis which corresponds to the area of the triangle whose magnitude is represented by the y-axis. Anything outside the domain will not form a triangle as the length given is too big or small.

Technically this should not be a polynomial however the question states that it is so I'm not sure what else to cover.

Regards.

 

[mmm4444bot]

Hello phob16:

Overall, your work looks good. I see a couple decisions need correcting and notation/wording issues; you misread the bit about 'polynomial', and your graph contains an error. Let's cover these issues.

… The magnitude of side length C …

c and C are two different symbols. Generally, we use uppercase letters for the triangle's angles (and/or vertices) and we use lowercase letters for side lengths. These rules aren't etched in stone (i.e., you're free to choose your own symbols, when they haven't been provided), but don't use two different symbols to represent the same thing.

… Anything outside the domain will not form a triangle as the length given is too big or small.

Very good. Knowing two sides (30 and 10), we can immediately say side c must be greater than 20 and less than 40. See this page, for the Triangle Inequality Theorem; they refer to it as, "Triangle inequality", so you may search the page for that name. Each side of a triangle must be less than the sum of the other two sides.

… I've managed to conclude that the zeros create a restriction for the domain that the side length [c] can exist as …

Yes, you correctly found the zeros of the functions: {-40, -20, 20, 40}. Throw away those negative values; they are not consistent with your use of g(c), as discussed below. They are extraneous information.

Your graph is plotted in Quadrants I and II. There are no negative side lengths, so there ought to be no graph in Quadrant II.

The Triangle Inequality Theorem tells you: 20 < c < 40. Any other value for the length of side c will not form a triangle. I don't see a need to include the origin, either. The relevant graph is over the domain of function g(c).

In your own words, what does the graph of g(c) m͏͏ean to you? Can you state function g's domain and range? I would include statements about each of those sets, in your description of the graph.

Technically this should not be a polynomial however the question states that it is …

You've misread that part. They're referring to the function f(c). It's the polynomial, not g(c). The area function g(c) models the triangle's area as the principal square root of a 4th-degree polynomial. (Note the word 'principal' -- we need the positive square root, to model area.)

Overall, I think you've done good work. You just need to tidy up some details. Let us know, if you have any more questions. Cheers :cool:

 

[mmm4444bot]

500 Infernal Server Error declines inclusion of the following text in post #2

I've managed to conclude that the zeros create a restriction for the domain [side length c]. The magnitude of [c] is [measured on the x-axis, and] the triangle [area is measured on the y-axis]

The edits (shown in square brackets) are for your consideration; they make the wording a little more precise. As an option, you're free to rename the axes. You could label the horizontal axis c and the veritcal axis g. If you decide to change the name g to A, be sure to update any labels on the graph. Or, instead of changing any names, you could state the following:

y = g(c)

x = c

The uploaded image of your work is fuzzy (original's dimensions too big, as noted in the forum guidelines); if you wrote A = f(c), that's incorrect. A=g(c). Function g depends on function f which in turn depends on c; g is a composition of the square-root function and f. (If you haven't studied composed functions, yet, you will!) Cheers