Graphing by Hand (calculus)

Graphing by Hand (calculus)


In this video, I ‘d like to show how to
graph by hand. In order to graph, you need a graphing paper,
a pencil, and calculus! Here are tips for graphing. First, you need to find x-intercepts and y-intercept,
if possible. For x-intercepts, setting y=0 to find x-values
and similarly, setting x=0 to find y-intercept. Then, investigate if there are any asymptotes
for fractional functions. For the horizontal asymptotes, we need to
take the limit at infinity. If you get the limit, then y=the limit value
should be the equation of horizontal asymptote. For the vertical asymptote, we need to set
the denominator function be zero to find the x-values and check the numerator function
if it is not zero when it takes that x-value. Once you checked, x=that x-value should
be the equation of vertical asymptote. Then, of course, find the behavior of increasing
or decreasing, and local maxima or minima of the function if any. Lastly, find out the concavity as well as
the inflection points. Those points guide you to graph easily. Let’s take a look at this example, F(x)=
x ^4th – 4x^3 First, Let y be zero to find the x-intercepts. Factor the right side of the equation by taking
the common factor x^3 . Then, Letting x ^3 and another factor (x-4) be zeros to get x
=0 or x=4. as x-intercepts. Next, letting x be zero to get y-intercept
as zero. Since this example is a polynomial function,
there are no asymptotes. Now investigate the behavior of increasing
or decreasing, and locate any local maxima or minima. First step, we need to find the critical values
which are possible maximum or minimum points. Taking the first derivative of the function
and letting it be zero to find the critical values. Taking the common factor 4x^2 and letting
each factor be zero to find the critical values, x=0 or 3. Step 2, making a table using the critical
values to see if the derivative is either positive or negative. Place two critical values on the lines and
make each interval using them. The first interval is x being less than the
smaller critical value, second interval is between the two critical values, and the last
one is greater than the larger critical value. Once we set up the intervals, choose arbitrary
test value for each interval to see the sign of the derivative. I chose -1 , 1, and 4 as the test values for
each interval respectively. Then we need to substitute those test values
into the derivative. You could use the derivative directly, however,
it is easier to use each factor of the derivative, 4x^2 and x -3. To find the sign of the derivative, we just
multiply the sign of each factor. For the first interval, substituting -1 in
to 4x^2 to get positive and -1 into x-3 to get negative. Multiplying positive by negative to get negative
for the first derivative. Continue to do the same way for the next interval. As I chose 1 as the test value, substituting
1 into 4x^2 and x-3 factors to get positive and negative respectively. Multiplying those signs to get negative for
the derivative. For the last interval, substituting the test
value 4 into 4x^2 and x-3 factors to get both positive, which gives the first derivative
positive as well. Now let’s find out the behavior of increasing
or decreasing. For the first and the second intervals, the
signs of the first derivative are negative, so the function is decreasing on those intervals. However, for the last interval, the sign of
the first derivative is positive, the function is increasing. Since the turning point from decreasing to
increasing should be the local minimum of the function, there is a minimum at x=3. Finally we need to find the minimum point
using the original function not the derivative function. Substituting x=3 into the original function
to get the local minimum point as (3, -27). Now, we found the local minimum at (3. -27). Lastly let’s find out the concavity of the
function as well as the inflection points. Let’s see the detailed work for that. The key for finding the inflection points
is that we need to use the second derivative of the function. First step, we need to find the possible inflection
points. Taking the second derivative of the function
and letting it be zero to find the values. Taking the common factor 12x and letting each
factor be zero to find the possible inflection values as 0 or 2. Step 2, making a table using the possible
inflection values to see if the second derivative is either positive or negative. Place two possible values on the lines and
make each interval using them. The first interval is x being less than the
smaller value, second interval is x being between the two values, and the last one is
x being greater than the larger value. Once we set up the intervals, choose arbitrary
test value for each interval to see the sign of the second derivative. I chose -1 , 1, and 3 as the test values for
each interval respectively. Then we need to substitute those test values
into the second derivative. You could use the second derivative directly,
however, it is easier to use each factor of the derivative, 12x and x -2. To find the sign of the second derivative,
we just multiply the sign of each factor. For the first interval, substituting -1 in
to12x to get negative. Again -1 into x-2 to get negative. Multiplying negative by negative to get positive
for the second derivative. Continue to the same way for the next interval. Substituting 1 into 12x and x-2 factors to
get positive and negative respectively. Multiplying those signs to get negative for
the second derivative. For the last interval, substituting the test
value 3 into 12x and x-2 factors to get both positive, which gives the second derivative
positive as well. Now let’s find out the concavity. For the first interval, the sign of the second
derivative is positive, so the function is concave up. For the second interval, the sign of the second
derivative is negative, so the function is concave down. For the last interval, the sign of the second
derivative is positive, so the function is concave up. At x=0, the concavity changes form
up to down, there is an inflection point. Similarly, at x=2, the concavity changes from
down to up, there is another inflection point. To find each inflection point, substitute
each x value to the original function. Substituting x=0 into the original function
to get the inflection point as (0, 0). And substituting x=2 in to the original function
to get another inflection point as (2, -16) Now, Let’s go back the previous slide. Lastly, we found the inflection points at
(0, 0) and (2, -16) Let’s put together the information of increasing,
decreasing, and concavity. The table 1 shows the behavior of increasing
and decreasing of the function and the 2nd table shows the concavity. Now we put two together to find the shape
of the graph in each interval. The important values are 0, 2, and 3, so we
use those values to create another table. In the same way as before, we set up each
interval using those three values. For the first interval, we can see the function
is decreasing from the first table and concave up from the second table. If we put them together, we can get the shape
as shown. For the second interval, the function is still
decreasing from the first table and concave down from the second table, so the shape is
like this. For the third interval, the function is still
decreasing but concave up, so the shape is like this. For the last interval, the function is now
increasing and concave up, so the shape is like this. Now let’s graph of the function based on
our investigations. We found x-intercepts, local minimum and two
inflection points. In addition, we found the shape of the graph
in each interval. So, Let’s graph the function. First, we need to set up the scales of x-and
y-axes based on the important values we found. For x-axis, it looks like 5 will be enough. As for y-axis, it will be fine as long as
we can plot -27. We do not have to set up the same scale for
x and y –axes. Next, we plot four important points which
we found, (0,0), (4, 0) as x-intercepts. Since (3, -27) is the local minimum, you could
draw a dashed guide line as shown. The last point is the inflection point (2,
-16). Now that we have all four important points,
all we have to do is drawing the shape for each interval based on the table we created. Finally, we did it! That’s is the graph of y=x^4 -4x^3 by hand.

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