Please follow the rules of posting in this forum, as enunciated at:
Let \(\displaystyle A: (-1,0),~B: (1,2),~C: (1,0),~\&~D: (3,0)\) then find:
Well Otis, please tell us what NikKen02, showed of hiis/her own work. I see nothing.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.
I already did that, pka. Here it is, again:Well Otis, please tell us what [NickKen02] showed …
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.… You began by writing:
g(x) = a · √[(x - h)^2] + k …
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.… Surely we all hope that the student will see the graph is a semicircle …
I posted that solving the equation… what you posted is wrong because \(\displaystyle y\) cannot be \(\displaystyle \pm\) …
Agree!… 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 …
Also agree! I don't understand why tutors would assume the exercise was created to have multiple answers.… the [exercise] author wanted the student to see a semi circle …
… I went through four ten-year reviews by regional accreditation groups …
It's been a long year!So much griping and snarkiness in this forum, lately. I hope next year is better.
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.)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)