Question

Let f(x)=a(x-p)^2+q be a quadratic function, where a,p,q are constants and a≠0. f(x) is translated up by k units, translated right by c units, and then take a reciprocal of the translated function. Define g(x) be the function after performing all the transformations.

a. Find the conditions on the constants a,q and k so that g(x) does not have a vertical asymptote.


b. Find the x value(s) so that there is/are invariant point(s) linking f(x) and 1/(f(x)).

Answer 1

translations give you

a((x-c)-p)^2+q+k
so,
g(x) = 1/(a(x-c-p)^2+q+k)
= 1/(a(x-(c+p))^2 + (q+k))
= 1/a(x^2-2(c+p)x+(c+p)^2 + (q+k))
= 1/(ax^2 - 2a(c+p)x + a(c+p)^2+q+k)

to have no vertical asymptote, the denominator cannot be zero, which means that the discriminant is negative.

4a^2(c+p)^2 - 4a(a(c+p)^2+q+k) < 0
4a^2(c+p)^2 - 4a^2(c+p)^2 - 4a(q+k) < 0
4a(q+k) > 0

By invariant, do you mean where f(x)=g(x)?

If so, then just set up the equation and solve for x. It'll be a quartic, so good luck there.

Answer 2

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