Help with a Function

Question

I'd like some help getting this question on its way, any help would be greatly appreciated.

For the following function

$$f(x) = \frac1{-(x-5)^2+7}\;,$$

determine:

a) The equation of the asymptote as exact values
b) Coordinates of the invariant points as exact values

Answer 1

a) This all depends on what asymptotes you are referring to. Let's work out each of the asymptotes.

Vertical asymptote

Note that there are no common factors in the numerator and denominator. Then, we can easily figure out the vertical asymptotes of that function. To do so, set 'zero' for the denominator expression and solve for the variable. That is:

$$-(x - 5)^2 + 7 = 0$$ $$7 = (x - 5)^2$$ $$\pm \sqrt{7} = x - 5$$ $$x = 5\pm \sqrt{7}$$

Those are the vertical asymptotes of the function.

Horizontal asymptote

The horizontal asymptote of the function occurs when there exists a real-valued constant $c$ such that

$$y = \lim_{x \rightarrow \infty} f(x) = c \text{ (where y is the variable)}$$

Suppose we want to determine $\lim_{x \rightarrow \infty} f(x)$. Then:

$$\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \dfrac{1}{-(x - 5)^2 + 7} = 0$$

So $y = 0$ which is the horizontal asymptote of the function.

Edit: Forgot to work out the second part.

b) I would assume that invariant points are points in which $f(c) = c$ where $c$ is the constant. As what Brian pointed out, you need to determine the values of $x$ to find the coordinates (though there aren't any such points.)

Answer 2

HINTS:

1. The vertical asymptotes, if any, occur at the values of $x$ where the denominator is $0$; to find them, solve the equation $-(x-5)^2+7=0$. The horizontal asymptotes, if any, are the lines $y=c$ and $y=d$, where $c=\lim\limits_{x\to-\infty}f(x)$ and $d=\lim\limits_{x\to\infty}f(x)$.

2. I’m assuming that an invariant point of $f$ is a number $a$ such that $f(a)=a$. To find those, you must solve the equation $f(x)=x$, i.e., the equation $$\frac1{-(x-5)^2+7}=x\;.$$ Unfortunately, this has no rational roots, so unless you’ve been taught the general solution of a cubic equation, I don’t see any way to get the exact values.

Added: If by invariant point you mean what I would call a critical point of the function, those occur where the function is defined and the derivative is $0$ or does not exist. Here the derivative turns out to be defined wherever the function is defined, so the critical points, if any, must be where $f\,'(x)=0$. To find them, just take the derivative of $f(x)$, set it to $0$, and solve for $x$. (If you’ve graphed the function, you should be able to make a good guess even before you do the calculation.)