SOLVE.
1 a. x^4-16=0 b. x^2=3 c. x^3=7
Sketch the following graphs.
2. a. f(x)=-(x-2)^2(x+3) b. (x-1)^2(x+3)(x+4)
$\displaystyle
f(x) = -(x-2)^2 (x+3)
$
$\displaystyle
f(x) = -(x-2)(x-2)(x+3)
$
so we know our roots are at $\displaystyle x = -3 $ and a double root at $\displaystyle x = 2 $
now using your knowledge of sketching graphs, a graph with 3 roots is going to have the shape of a sideways "S"
create a number line with these roots and plug in values in those intervals to see the exact shape of the graph. then sketch it
you try the next one
Yes, I just dont know when to draw the curves on the graph since I just switched into this class. What about the first question? Do you think you can send me a website on how to draw the curves on the graph when its cubic and quartic/ and positive/negative.
ok for the number line you have:
< ------(-3)----------(+2)-------->
take the equation we just created $\displaystyle f(x) = -(x-2)(x-2)(x+3) $ and plug in 3 different values from these intervals
use the numbers (-4) , (0) , and (3) in this equation
<---..(+)..----(-3)----..(-)..-----(+2)----..(-)..---->
so now you know how the graph looks.
the graph is above the x axis until it meets the root of (-3)
then the graph is below the x axis until it meets the root of (+2) but where this is a double root, the axis then continues back down below the x axis cause a sideways "s" shape