Notice that f(x)= 1x is defined for all real numbers except x=0. Therefore, there is no y-intercept. Moreover, since the numerator is never 0, there is no x-intercept. Knowing this, let's connect the points with a smooth curve.
b
We will examine the graph of g(x)= 4x+2 by comparing the graph of f(x)= 1x. To do so, let's remember the general form of the reciprocal function family.
y=a/x- h+ kIn this form, the value of a refers to stretches, shrinks, or reflections of the parent function f(x)= 1x. Moreover, the value of h represents the horizontal translation and k represents the vertical translation. With this in mind, let's examine g(x).
g(x) = & 4/x+2
⇕
g(x) = & 4/x-( -2)+ 0
As we can see, g(x) is a stretch of the graph of f(x) by a factor of 4 and it is followed by a translation 2 units to the left. Let's see them on the same coordinate plane.
c
We are asked to find the horizontal asymptote of the graph of g(x). Let's remember that in the general form of the reciprocal function, y= ax- h+ k, the line y= k represents the horizontal asymptote.
g(x) = & 4/x-( -2)+ 0
Therefore, y= 0 is the horizontal asymptote of g(x).
d
This time we will identify the vertical asymptote. One more time, we will recall that in the general form of the reciprocal function y= ax- h+ k, the line x= h represents the vertical asymptote.
g(x) = & 4/x-( -2)+ 0
Therefore, x= -2 is the vertical asymptote of g(x). Notice that the function is undefined at x= -2 because it makes the denominator 0.
x-( -2)=0 ⇔ x= -2
This means that -2 is not included in the domain of the function. We can conclude that the domain of a rational function does not include the vertical asymptotes.
