so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.
in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.
now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B7%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B%5Cfrac%7B19%7D%7B2%7D%7D~%2C~%5Cstackrel%7By_2%7D%7B%5Cfrac%7B7%7D%7B2%7D%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D-7%20%5Cright%29%5E2%2B%5Cleft%28%20%5Cfrac%7B7%7D%7B2%7D-4%20%5Cright%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B5%7D%7B2%7D%5Cright%29%5E2%2B%5Cleft%28%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%7D%5Cimplies%20%5Cboxed%7BAM%5Capprox%202.549509756796392%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
Analyzed and Sketched.
Step-by-step explanation:
We are given 
To sketch the graph we need to find 2 components.
1) First derivative of y with respect to x to determine the interval where function increases and decreases.
2) Second derivative of y with respect to x to determine the interval where function is concave up and concave down.
is absolute maximum
is the point concavity changes from down to up.
Here, x = 0 is vertical asymptote and y = 0 is horizontal asymptote.
The graph is given in the attachment.
The domain is the origin of the arrow
domain = {-2,2,3,4}
Option d is the answer
Answer:2.87
Step-by-step explanation:
The answer to ya question is 12.88
