Give an example of a function with both a removable and a non-removable discontinuity.

(x+5) (x=3)

(X+5) (x+1)

Step-by-step explanation:

A removeable discontinuity is always found in the denominator of a rational function and is one that can be reduced away with an identical term in the numerator. It is still, however, a problem because it causes the denominator to equal 0 if filled in with the necessary value of x. In my function above, the terms (x + 5) in the numerator and denominator can cancel each other out, leaving a hole in your graph at -5 since x doesn't exist at -5, but the x + 1 doesn't have anything to cancel out with, so this will present as a vertical asymptote in your graph at x = -1, a nonremoveable discontinuity.

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)