F (X) 2/x^2-16

Question

Find the

X= intercept

y= intercept

max

min

Vertical

horizonal

holes

Answer 1

In this problem to find the x and y intercept is using previous knowledge. For and function the x intercept is the the values of F(x) when x = 0. So plug zero in for x and solve for F(x). The y-intercept is the x-values for when F(x)=0. So make F(x) = 0 and solve for x. In this problem this will require you to muliply by the denominator and actually obtain the answer 0 = 2 which is impossible so there are no x intercepts. the y-intercepts are -1/8. The vertical asymptote is what makes your denominator zero. so I set the denominator equal to zero and solve. By using the difference of two squares I find (x-4)(x+4) = 0 where the vertical and horizontal asymptoes are located at x = 4 and -4. To find the horizontal asymptotes you need to look at the degrees of the polynomials of the numerator and denominator. Since the numberator is a constant the degree is 0 which is less than 2. By using the definition located in your book or on the internet. The horizontal asymptote is y = 0. Since I cannot factor my rational expression to obtain a linear equation there are no holes in my graph. I have presented plenty of videos on how to do these types of problems, you may also get an explanation on these on my youtube channel located at http://www.youtube.com/mrbrianmclogan Let me know if it helps!