SOLUTION: Find the y-intercept and graphing of this rational function? (8x-4x^2)/((x+2)^2) VA: x =-2 HA: y = -4 X-int: (2,0),(0,0) How would I find the y-intercept How would I

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Question 916869: Find the y-intercept and graphing of this rational function?
(8x-4x^2)/((x+2)^2)
VA: x =-2
HA: y = -4
X-int: (2,0),(0,0)
How would I find the y-intercept
How would I go about graphing this as well?
Thank you!
Found 2 solutions by solver91311, Theo:
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

The -intercept is the point where is the value of the function when .

Plot the vertical asymptote and the horizontal asymptote. Plot the three intercepts.

Use the two -intercepts and the intersection of the vertical asymptote to divide the -axis into four regions, namely the three intervals , ,, and . Select a value from each of the intervals and evaluate the function at that value. The sign of the function value will tell you whether the graph in that region is above or below the -axis. The graph will tend to infinity (positive or negative depending on the sign of the function in that region as determined above). As increases or decreases without bound, the graph will tend to the horizontal asymptote. And the graph will pass through each of the three intercepts that you have plotted. A rough sketch with these parameters should suffice for your needs.

John

My calculator said it, I believe it, that settles it

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the y-intercept is the value of y when x = 0.
your expression is:
(8x-4x^2) / (x+2)^2
set that equal to y and you get:
y = (8x-4x^2) / (x+2)^2
set x = 0 and you get:
y = 0 / 4 = 0
your y-intercept is y = 0
that's the value of y when x is equal to 0.

to manually graph this, plot the points that you know about.

you know that as the value of x goes to infinity, the value of y will approach -4 but be exactly 4.
you know that as the value of x goes to minus infinity, the value of y will approach -4 but never be exactly 4.
with horizontal asymptotes, the graph of the equation can cross the horizontal asymptote but it will never stay there. it will either be slightly above it or slightly below it. the further out you go, the closer it gets to -4 but will always not be right on it.

so draw a horizontal line at y = -4 and that will be your guide for the horizontal asymptote.

your vertical asymptote will be at x = -2 so draw a vertical line at x = -2.
that will bhe your guide for the vertical asymptote.

plot your y-intercept point at (0,0) and plot your x-intercept point at (0,0).

your graph will have to go through that point.

now plot some points around these landmarks.

plot a point at x = -3 and x = -1.
plot another point at x = -2.5 and x = -1.5
that will tell you whether the graph is approaching minus infinity or positive infinity on which side of the vertical asymptote.

plot some points at x = 10 and 100 and -10 and -100.
that will tell you how fast the graph is approaching the horizontal asymptote.

since the graph crosses the x-axis at (0,0) then, if it goes positive after (0,0), it will have to go back negative at some point in time.

you have the x-intercepts at 0 and 2 so it either goes negative at 0 and then turns positive at 2 or it goes positive at x = 0 and turns negative at x = 2
since the horizontal asymptote is -4, it must be going positive at x = 0 and then going back negative at x = 2.

all these hints give yoou an idea about how the graph will look.

you can cheat like i do by using a graphing calculator to see how the graph will end up. it's a good idea to try to do it manually and then look at the graph to see how you did and where you went right and where you went wrong.

i graphed it using a graphing calculator and the following is what i got.