Jessica Carlson
Group Homework
Sarah Raubinger
10/19/06
Christie Donahue
56. Find the lines through the origin that are tangent to the parabola.
y=x
2
2x+4
X
Y
3
19
2
12
1
7
0
4
1
3
2
4
3
7
The derivative of the parabola is the slope of the tangent line at a given point.
Using the
Power Rule (d/dx(x
n
)=nx
n1
), we can find the equation of y’ by bringing down the power of
2 down in front of the x in the first term of the equation making the term 2x.
The second
term will become a 2 since the power on the x is a 1.
The last term will drop out since
there is the “power on the x” is a 0.
y’=2x2
As you can see in the graph above, two lines (one on each side of the yaxis) exist that are
line tangent to the graph and that pass through the point (0,0).
Since we know the slope of the line is equal to the derivative of the original equation at a
given point
, we can use the Slope Formula (y
2
y
1
)/(x
2
x
1
) to find the slope of a line that is
tangent to the graph of the original function AND passes through the point (0,0).
Since we
don’t know the exact point on the graph that the tangent will pass through, we can use the
an arbitrary
values
of x and y for the graph of the function (x, x
2
2x+4).
Thus, our two
points for the slope formula are (x, x
2
2x+4) and (0,0).
We used these points as (x
1
,y
1
) and
(x
2
,y
2
) respectively (although they could be switched to obtain the same answer).
These
two points in the slope formula are equal to the derivative of the equation of the graph,
which we found above.
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 Fall '08
 BLAKELOCK
 Calculus, Derivative, Slope, Group Homework, Jessica Carlson Sarah Raubinger Christie Donahue, Christie Donahue, Jessica Carlson, Sarah Raubinger

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