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Good day professor

I've manage to do most of it but just got stuck with 2.5 and 2.6! So the assistance would be appreciate it! Many thanks :)

I've attached a link to the question and a picture!

http://rs1287.pbsrc.com/albums/a626/Chantell_Young/Mobile%20Uploads/utf-8BcXVlc3

For Probs 2.5 and 2.6, we can assume you have already solved for the constants a, b, c for the parabola f(x) and m and y0 for the line g(x) given by

g(x) = mx + y0.

We can also assume that you have figured out the coordinate of the midpoint of g(x), i.e., (xm, ym)

For 2.5, note that a hyperbola has y ~ 1/x and in general can be written

h(x) = A/(x-x∞)

where A is an overall scaling factor and x∞ is the value of x where the hyperbola goes off to infinity. Since we have 2 unknowns, A and x∞, but only one eqn between them, namely h(xm) = ym, coupled with the fact that xm < 0 so that x∞ cannot be 0, it looks lie we need to assume a value for A. So let A = 1.

So now just set

h(xm) = ym = 1/(xm - x∞)

and solve for x∞.

For 2.6, we need to find the maximum of a function z(x) = distance between f and g. Finding an extremum, such as a maximum, involves setting the derivative equal to zero, so first find the eqn for z(x) and take the derivative. So

z(x) = ax^2 + bx + c - mx - y0

and

dz/dx = 2ax + b - m = 0

and solve for x. ( I get x = (m - b)/2a ).

Good luck!

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