sketch

• February 22nd 2010, 04:05 AM
furor celtica
sketch
i am told in my book, in the middle of a domain/range section:
sketch the graph of x(8-2x)(22-2x)
i really have no idea of how to do this, any tips?
• February 22nd 2010, 07:20 AM
NOX Andrew
First, determine the domain and range of the graph. In this case, both the domain and range are all real numbers.

Next, determine the intercepts, asymptotes, and symmetry of the graph.

Start by finding any x-intercepts.

$x(8-2x)(22-2x) = 0$
$x = 0$ or $8 - 2x = 0$ or $22 - 2x = 0$
$x = 0$ or $x = 4$ or $x = 11$
$x = \{0, 4, 11\}$

Now find any y-intercepts.

$y = 0(8 - 2 \times 0)(22 - 2 \times 0)$
$y = 0(8 - 0)(22 - 0)$
$y = 0(8)(22)$
$y = 0$

There aren't any vertical asymptotes (because there isn't division by zero, square roots, etc.). There aren't any horizontal asymptotes either because, as x increases or decreases without bound, the graph decreases without bound.

Now determine if the graph is symmetric with respect to the y-axis.

$-x(8 - 2(-x))(22 - 2(-x)) = -x(8 + 2x)(22 + 2x)$

Because -x(8 + 2x)(22 + 2x) is not equal to x(8 - 2x)(22 - 2x), the graph is not symmetric with respect to the y-axis.

Now let y = x(8 - 2x)(22 - 2x) to determine if the graph is symmetric with respect to the x-axis.

$-y = x(8 - 2x)(22 - 2x)$
$y = -x(8 - 2x)(22 - 2x)$

Because -x(8 - 2x)(22 - 2x) is not equal to x(8 - 2x)(22 - 2x), the graph is not symmetric with respect to the x-axis.

I have to go to class but I'll continue if nobody else has answered your question by the time I get home.
• February 22nd 2010, 07:23 AM
HallsofIvy
Quote:

Originally Posted by furor celtica
i am told in my book, in the middle of a domain/range section:
sketch the graph of x(8-2x)(22-2x)
i really have no idea of how to do this, any tips?

There are no fractions or anything there? Well, you can simplify that by taking a factor of 4 out of each of 8- 2x= 2(4- x) and 22- 2x= 2(11- x) to get
y= 4x(4- x)(11- x). Now you should be able to see that y= 0 if x= 0, 4, or 11 so you can mark (0, 0), (4, 0) and (11, 0) on your graph. when x< 0 , both 11-x and 4- x will be positive so 4x(4-x)(11-x) is the product of 3 positive numbers and one negative number. That tells you that y is negative for x< 0. Of course one of those factors will change sign at x= 0, x= 4, and x= 11 so the sign of y will alternate between negative and positive.

That is, y< 0 for x< 0, y> 0 for 0< x< 4, y< 0 for 4< x< 11, and y> 0 for x> 11. The graph will be below the x-axis for x< 0, above the x-axis for 0< x< 4, below the x-axis for 4< x< 11, and above the x-axis for x> 11.
y will go rapidly toward negative infinity as x< 0 decreases anmd will go to positive infinity for x>11 increases.

That should be enough to make a rough sketch. To be more precise you would need to calculate specific y values for a number of different x values.