I need help w/ these types of radical funcs

Hello NickKen02. Thanks for showing your work so far. You began by writing:

g(x) = a · √[(x - h)^2] + k

That equation doesn't match the equation for a semi-circle, so I'm wondering whether you've realized that the given graph shows the upper half of a circle. (The plotted semi-circle appears a bit squashed because the vertical scale wasn't set the same as the horizontal scale -- it's been compressed.)

Here's the standard equation for a circle with radius r, centered at coordinates (h,k):

(x - h)^2 + (y - k)^2 = r^2

You can get the values of h, k and r directly from the graph, by using the labeled coordinates. Substitute those values for h, k and r into the standard equation, and then solve for y.

The results will look like y = ±√[f(x)], where symbol y represents g(x) and symbol f(x) is a quadratic polynomial. Of course, you ignore the result with the negative sign in front because that corresponds to the lower half of the circle.

If you get stuck, please show your attempt.

?
 
Otis said:
Hello NickKen02. Thanks for showing your work so far. You began by writing:
g(x) = a · √[(x - h)^2] + k
That equation doesn't match the equation for a semi-circle, so I'm wondering whether you've realized that the given graph shows the upper half of a circle. (The plotted semi-circle appears a bit squashed because the vertical scale wasn't set the same as the horizontal scale -- it's been compressed.)
Here's the standard equation for a circle with radius r, centered at coordinates (h,k):
(x - h)^2 + (y - k)^2 = r^2
You can get the values of h, k and r directly from the graph, by using the labeled coordinates. Substitute those values for h, k and r into the standard equation, and then solve for y.
The results will look like y = ±√[f(x)], where symbol y represents g(x) and symbol f(x) is a quadratic polynomial. Of course, you ignore the result with the negative sign in front because that corresponds to the lower half of the circle.
Click to expand...
Well Otis, please tell us what NikKen02, showed of hiis/her own work. I see nothing.
Surely we all hope that the student will see the graph is a semicircle.
BTW what you posted is wrong because \(\displaystyle y\) cannot be \(\displaystyle \pm\), do you even see why?
 
I do not see why the graph is necessarily a semi-circle. We are given three points, none below the x-axis. It is reasonable to assume from the graph that one of those points is a local maximum, that the function is symmetric around the axis of x = 1, that the function has no points of inflection in its domain, and that the function is increasing (decreasing) at a decreasing (increasing) rate to the left (right) of the axis of symmetry. Obviously, a semi-circle with radius of 2 meets those criteria, but do they entail a semi-circle?
 
To supplement what I wrote in my previous post, it is given or implied that g(x) is differentiable and

[MATH]g(x) = \sqrt{f(x)},\ g(1 - z) = g(1 + z) \text { if } -\ 2 \le z \le 2, g(3) = 0,\ g(1) = 2,[/MATH]
[MATH]g'(x) > 0 \text { if } -\ 1 < x < 1, g'(1) = 0, \text { and } g''(x) < 0 \text { if } -\ 1 < x < 3.[/MATH]
We are to identify f(x).

You can prove that f(x) may represent a unique ellipse, which happens to be a circle.

But it is easy to generate other functions that satisfy the conditions. For example

[MATH]f(x) = (1.5 + x - 0.5x^2)^2 \implies g(x) = 1.5 + x - 0.5x^2 \implies g'(x) = 1 - x \implies g''(x) = -\ 1.[/MATH]
Now it is obvious by inspection that g'(x) > 0 if - 1 < x < 1, that g'(1) = 0, and that g''(x) < 0 everywhere.

g(3) = 1.5 + 3 - (3^2/2) = 4.5 - 9/2 = 0.

g(-1) = 1.5 - 1 -((-1)^2) = 0.5 - 1/2 = 0.

g(1) = 1.5 + 1 - (1^2/2) = 2.5 - 0.5 = 2.

Finally, for symmetry

[MATH]g(1 - z) = 1.5 + (1 - z) - 0.5(1 - z)^2 = 2.5 - z - 0.5( 1 - 2z + z^2) = 2 - 0.5z^2.[/MATH]
[MATH]g(1 + z) = 1.5 + (1 + z) - 0.5(1 + z)^2 = 2.5 + z - 0.5(1 + 2z + z^2) = 2 - 0.5z^2.[/MATH]
In fact, the square of any function that is symmetric around x = 1 and contains the three specified points is a valid answer. Bad problem.
 
I think we are making the solution complicated - uneccesarily.

3 points given on the curve (with 2 x-intercepts). If I were to do this problem, I would assume:

y = A* (x+1) * (x-3)

Curve passes through (1,2). So:

2 = A * 2 * (-2) \(\displaystyle \ \to \ \ \) A = -1/2

Thus:

g(x) = (-1/2) * (x + 1) * (x - 3)

According to OP the problem is solved here - there is no mention of symmetry and there was no need to calculate f(x).

Of course, there could be many more functions that could be passing through these points.
 
I disagree. I think that whoever wrote the question meant it for the usual algebra student and thereby expected that the student to see a semicircle centered at \(\displaystyle (1,0)\) and radius \(\displaystyle 2\). The whole circle will be \(\displaystyle (x-1)^2+y^2=4\) so that
the semicircle is \(\displaystyle y=\sqrt{4-(x-1)^2}=\sqrt{3+2x-x^2}\).
Thus \(\displaystyle f(x)=3+2x-x^2\) Then the semicircle is \(\displaystyle g(x)=\sqrt{f(x)}~.\)
 
A couple of comments here. First the student posted this problem in the Algebra section so using calculus or referring to precalculus might be the wrong path to take with this student. I see what Khan is saying but I agree with pka that the author wanted the student to see a semi circle. I think that is the way to proceed with this problem and if one likes they can mention that many other functions contain those three points and actually can give some of those functions. Personally I feel that assuming the graph is a semicircle will be the easiest way to get a function that passes the three given points AND looks likes the given graph.
 
pka said:
Well Otis, please tell us what [NickKen02] showed …
Click to expand...
I already did that, pka. Here it is, again:

Otis said:
… You began by writing:

g(x) = a · √[(x - h)^2] + k …
Click to expand...
If people are unable to see the handwritten work shown at the bottom of the op image, perhaps their device is not displaying the entire file.

pka said:
… Surely we all hope that the student will see the graph is a semicircle …
Click to expand...
Of course, but we don't exist in an ideal world. My approach is to confirm basics at the beginning of the conversation, whenever my instincts warn me that something basic may be amiss.

… what you posted is wrong because \(\displaystyle y\) cannot be \(\displaystyle \pm\) …
Click to expand...
I posted that solving the equation

(x - h)^2 + (y - k)^2 = r^2

for y results in two radical expressions and that we need to ignore the negative one . I didn't realize that people would read my statements to mean that both expressions comprise the final answer.

\[\;\]
 
Well I am know that the descriptions of algebra, college algebra, and pre-calculus are all over the board.
They are either disjoint or overlapping. I went through four ten-year reviews by regional accreditation groups. Every so called teams had different ideas as to what terms mean. It ain't a pleasant experience. All of that to say that I place no conference in where something is posted.
 
I agree that the student may be meant to assume it's either an ellipse or a circle (circularity seems to be deliberately hidden); but I'm troubled by the idea of telling them that they are "determining the equation algebraically", as if the answer was certain. Guessing is appropriate in real life, but it ought to be made clear when that is being done. If we object to telling students to "determine the formula" for a sequence given a few terms, this is no better.

We don't know (and apparently never will) the context of the problem. Have they learned about equations of ellipses or not? From what we can see, the topic is radicals, not conics, and the most natural guess might be that f(x) is a polynomial (which the OP appears to have assumed, but then placed the radical incorrectly). That is a different guess that would lead to the same answer, done correctly.

In fact, that was my first response to the problem: Assume f(x) is a quadratic with vertex (1,4) and intercepts (-1,0) and (3,0); it turns out to be [MATH]f(x) = -(x+1)(x-3) = -x^2 + 2x + 3[/MATH], and [MATH]g(x) = \sqrt{-x^2 + 2x + 3}[/MATH].
 
Jomo said:
… First the student posted this problem in the Algebra section so using calculus or referring to precalculus might be the wrong path to take with this student …
Click to expand...
Agree!

… the [exercise] author wanted the student to see a semi circle …
Click to expand...
Also agree! I don't understand why tutors would assume the exercise was created to have multiple answers.

As far as f(x) goes, the exercise clearly states that the graph represents a function g that's defined as the square root of a function f. The exercise asks for that definition.

What's wrong with providing the answer in the given form?

So much griping and snarkiness in this forum, lately. I hope next year is better.

:(
 
The question to answer in OP (the visible part) was:

  • Determine, algebraically, the equation for g(x)....... There might be other "intended" part to this problem.
 
Subhotosh Khan said:
3 points given on the curve (with 2 x-intercepts). If I were to do this problem, I would assume:

y = A* (x+1) * (x-3)

Curve passes through (1,2). So:

2 = A * 2 * (-2) \(\displaystyle \ \to \ \ \) A = -1/2

Thus:

g(x) = (-1/2) * (x + 1) * (x - 3)
Click to expand...
If we could assume that the graph shown represents a polynomial, then this would be valid; but the vertical slopes at the intercepts make it clear (to someone with deeper experience than introductory algebra) that it is not a polynomial; and you have ignored the fact that g is said to be the square root of some function f, so that g is not assumed to be a polynomial. (Okay, your function could be written as [MATH]g(x) = \sqrt{\frac{1}{4}(x + 1)^2(x - 3)^2}[/MATH], but it still wouldn't really have that graph.)

I still say it is a poorly written problem.
 
I agree with Dr. Peterson that it is a badly written problem. Nor do I think that I was being snarky toward any tutor or student.

Obviously, the problem said to solve algebraically. And I recognize that my work on the problem used calculus, but that was not addressed to the student. Moreover, I looked at the graph and the relation between the possible functions, which is also non-algebraic thinking, in part because it seems plausible to me that the student was expected to use the graph.

If the intention of the problem was to find A function that goes through the three given points, there are an infinite number of functions that will do so. Teaching a student that a function can be determined by looking at three points is not merely useless, it is positively deleterious. We try to train everyone in mathematics in order to teach people rigorous thinking. Making definitive conclusions based on scanty data is exactly what we should not be teaching.

If, as I suspect, the purpose was to find THE function that goes through the three points and has other characteristics that were to be deduced from the graph, those characteristics (such as the absence of inflection points) cannot be determined for certain from a graph and are difficult to handle except in the language of calculus. (And there is the oddity that the graph was scaled in an unnecessarily deceptive way.)

In short, I continue to think it is a horrible problem and grossly unfair to the student. We have a bunch of tutors arguing about what the problem means; think about what that implies for the student subjected to this text.
 
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