Monday

September 26, 2016
Posted by **alicia** on Tuesday, February 26, 2008 at 5:44pm.

y= x(x+3)^2(x+5)

y=x(x-2)(x+0.5)(2x-1)

- algerbra 2 -
**Damon**, Tuesday, February 26, 2008 at 6:31pmy= x(x+3)^2(x+5)

I am not going to tell you how to graph that, but I will tell you how to sketch the graph.

for very large positive x, the thing is large positive, so in the end it will slope up to the right. So over on the right in the first quadrant draw an arrow pointing up to the right.

for very large negative x, because the highest power of this polynomial is x^4 which is even, y will again be large positive. So over in the left quadrant you have an arrow pointing up to the left.

now it is zero at x = 0 , x - -3 (twice), x = -5

Mark those 3 or really four but two are the same spots with dots on your graph.

SO

Your function comes screaming down from your arrow at the upper left and passes through (-5,0) headed down.

then it turns up and grazes the x axis again at (-3,0). I say grazes because it did not have enough oomph to get through the x axis on its way up and just touched it, heading back down. If it had gone through, we would have had two zeros close to x = -3, but they are scrunched together because the function just touched the axis there and dropped back down.

then it heads back up again to pass through the origin (0,0) on its way up to your arrow at the upper right.

Now you should be able to do the second one. - algerbra 2 -
**Reiny**, Tuesday, February 26, 2008 at 6:39pmto solve, let y = 0

since the function is already factored, this is now easy

x(x+3)^2(x+5) = 0

so x=0 or x=-3 or x=-5

to graph it, realize that the solution to the corresponding equation is the x-intercept of the function.

When you have a "double root" such as you get from (x+3)^2, the graph will touch the x-axis without crossing over.

Furthermore, since the highest term is +x^4 the curve will "get lost" up in the first and second quadrants.

So looking at our solutions, the graph will cross at -5, touch at -3 and cross again at 0

I then substitute a value of x about 1/2 between x-inercepts to see approximately how high or how low the graph goes between x-intercepts.

at x=-4, y = -4

at x=-1, y = -15

at x=-2, y = -18

This gives me a pretty good idea about the behaviour of my graph.

Using the same type of logic you should be able to do the next one.