group hw 5 - Jessica Carlson Sarah Raubinger Christie...

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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 n-1 ), 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’=2x-2 As you can see in the graph above, two lines (one on each side of the y-axis) 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. 0-(x
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This note was uploaded on 04/02/2008 for the course MATH 115 taught by Professor Blakelock during the Fall '08 term at University of Michigan.

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group hw 5 - Jessica Carlson Sarah Raubinger Christie...

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