I need to graph this function but don't know if I should use critcal points to grpah this or just asymptotes?

Question

I have a function that's (x+4) / (x-3)

Here, it's not neccesary for me to find the concavity point or even the critical points since I can find the asymptotes right...if there was not a vertical asymptite then I would have to find all the critical and second derivative right? I just want to make sure of what exactly I should find in this function and if I'm right of whatever I said, Please clarify this for me. Thank You.

Answer 1

Hi,

Your graph will have a vertical asymptote at x = 3 because of the x - 3 expression in the denominator.

Because the top and bottom expressions are both of the first degree, there is a horizontal asymptote at y = 1 because of
==>1x + 4
.......---------
==>1x - 3

The graph will be in the upper right hand corner and the lower left hand corner of the sections formed by the asymptotes.

You can get specific points by plugging in x values or by putting the equation in a graphing calculator and look at the table.

I hope that helps!! :-)

Answer 2

First off, critical points are defined to be where f'(x) = 0 or where f'(x) is undefined.

Also, critical points tell us where intervals of increase/decrease as well as intervals of concavity may alternate. This is clarified in my last post. Complete solution here:

f(x) = (x + 4)/(x - 3)

Solving for f'(x) using the quotient rule,

f'(x) = [(1)(x - 3) - (x + 4)(1)] / (x - 3)^2
f'(x) = [x - 3 - x - 4] / (x - 3)^2
f'(x) = -7/(x - 3)^2

Critical points are where f'(x) is equal to 0 _OR_ where f'(x) is undefined. In this case, f'(x) is undefined when the denominator is equal to 0.

(x - 3)^2 = 0
x - 3 = 0
x = 3

Critical point: x = 3.

To determine the intervals of increase/decrease, test a single point to the left of x = 3 and to the right of x = 3, for f'(x). We want to test if it is positive/negative. First, draw a number lne containing the critical points.

. . . . . . . . . (3) . . . . . . . . . . . .

If x = 2, then f'(2) = -7/(2 - 3)^2 = -7/(-1)^2 = -7/1 = -7, which is negative. Mark the region as negative.

. . . [-] . . . . (3) . . . . . . . . . . . .

If x = 4, then f'(4) = -7/(4 - 3)^2 = -7. Mark the region as negative.

. . . [-] . . . . (3) . . . . . .[-]. . . . .

In the negative regions, the function is decreasing. But, for critical points, it is not included in the interval.

f(x) is decreasing from (-infinity, 3) U (3, infinity).
f(x) is increasing nowhere.

Now, let's find the second derivative, to obtain our intervals for concavity. Recall that

f'(x) = -7/(x - 3)^2

Skipping the details,

f''(x) = 14/(x - 3)^3

Again, draw a number line to test f''(x) for positivity/negativity at our critical points. In this case, our only critical point is x = 3.

. . . . . . . . . . (3) . . . . . . . . . .

Test x = 2: f''(2) = 14/(2 - 3)^3 = 14/(-1)^3 = -14, which is negative.

. . . . {-} . . . . (3) . . . . . . . . . .

Test x = 4. f''(4) = 14/(4 - 3)^3 = 14/1^3 = 14, which is positive.

. . . . {-} . . . . (3) . . . .{+}. . . . .

Our function is not defined at our critical point x = 3, so we won't have local extrema.

f(x) is concave down on (-infinity, 3).
f(x) is concave up on (3, infinity).

There is no inflection point, because the function is not defined at x = 3.

Answer 3

The y-asymptote is y=1 and the x-asymptote is x=3.
But you should know that the function has a root at x =-4 so that you can accurately plot that critical point.

Answer 4

Easy if you just consider this is 1 + 7/(x-3), essentially the form of 1/x with a couple shifts and one magnification.

Answer 5

You should find the apical points of this hyperbola and its asymptotes that's should be enough.