I assume you don't need to find any maxima, minima, or inflection
points here. I assume also that you don't need Calculus here.
Because the y' and y'' are not that simple for this rational function.
I assume you only need to sketch the graph.
-----------
y = (2x^2 -3)/(x+2)
Step 1. Find any vertical asymptote.
The roots of the denominator are the vertical asymptotes.
x+2 = 0
x = -2 ........the only vertical asymptote.
------------
Step 2. Find any horizontal asymptote.
If there are limits for y as x approaches (- infinity) or (+
infinity), then those limits are the horizontal asymptotes.
Note:
>>> n /(- infinity) = -(n)/(-(-infinity)) = -n/(+infinity) = zero, almost.
>>> n /(+infinity) = zero, almost.
Meaning, any non-zero number divided by +,-infinity will tend/approach zero.
y = (2x^2 -3)/(x +2)
We divide both numerator and denominator by x^2,
y = [(2x^2)/(x^2) -3/(x^2)] / [x/(x^2) +2/(x^2)]
y = [2 -3/(x^2)] / [1/x +2/(x^2)]
When x approaches +,-infinity,
y = [2 -0] / [ 0 +0] = 2/0 = +,-infinity almost.
That means there are no horizontal asymptotes.
------------------
Step 3. Find other asymptotes.
y = (2x^2 -3)/(x+2)
Since the degree of the numerator, 2, is one degree higher than the
degree of the denominator, 1, then there is a linear oblique
asymptote.
We can find that by long division.
y = 2x -4, remainder 5/(x+2) ----by long division.
That means, y = 2x -4 is the oblique asymptote.
(Have this been taught to you yet?)
---------
Step 4. Find x-intercepts.
At x-intercepts, y=0, so,
0 = (2x^2 -3)/(x+2)
Multiply both sides by x+2,
0 = 2x^2 -3
x^2 = 3/2
x = +,-sqrt(3/2) = +,-1.225 approx.
Therefore, points (-1.225,0) and (1.225,0) are the two x-intercepts.
---------
Step 5. Find y-intercepts.
At y-intercepts, x=0, so,
y = (2(0)^2 -3)/(0 +2)
y = -3/2 = -1.5
Therefore, point (0,-1.5) is the only y-intercept.
----------
Step 6. Investigate the intervals of x that are subdivided by the
vertical and oblique asymptotes.
,,,,,,
In the interval (-infinity,-2). Or, from -infinity to close to -2,
Say, when x = -4,
y = (2(-4)^2 -3)/(-4+2) = 29 / -2 = -14.5
When x = -5,
y = (2(-5)^2 -3)/(-5 +2) = 47 / -3 = -15.67 ....lower than -14.5 at
y= -4. (...just as I thought.)
When x = -3,
y = (2(-3)^2 -3)/(-3 +2) = 15 / -1 = -15 ....lower also than -14.5 at
y= -4. (...yes.)
,,,,,,,
In the interval (-2,2]. Or, from just after -2 to 2.
(By the way, the oblique asymptote y = 2x -4 intersects the x-axis at
x=2. Because at the x-axis, y=0, so, 0 = 2x -4. So, x=2.)
At x before the left x-intercept. Say at x = -1.5,
y = (2(-1.5)^2 -3)/(-1.5 +2) = 1.5 / 0.5 = 3 ....above the x-axis.
(...just as I thought, ...due to the vertical asymptote.)
At x after the right x-intercept. Say at x = 1.5,
y = (2(1.5)^2 -3)/(1.5 +2) = 1.5 / 3.5 = 0.43....above the x-axis also. (...yes.)
-------------------
There, you have enough points to sketch the graph of y = (2x^2 -3) / (x+2).
Draw the x,y axes,
Draw the two asymptotes,
Plot the x- and y-intercepts,
Plot the extra points (-5,-15.67), (-4,-14.5), (-3,-15), (-1.5,3), and (1.5,0.43).
Connect the points, and extend the two branches of the graph up or
down towards the asymptotes.
The graph has two branches---one opening downwards, and the other, upwards.
It looks like an "oblique" hyperbola. |
