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May 29, 2015

May 29, 2015

Posted by **mehak** on Sunday, October 30, 2011 at 9:38pm.

- calculus -
**Steve**, Monday, October 31, 2011 at 12:21pmy = 23 - x

^{2}

y' = -2x

So, we want all points where the line through (x,y) and (10,26) has slope -2x

(y-26)/(x-10) = -2x

(23-x^{2}- 26)/(x-10) = -2x

-3 - x^{2}= -2x(x-10)

-3 - x^{2}= -2x^{2}+ 20x

x^{2}- 20x - 3 = 0

x = -0.149 or 20.149

At those values for x, we have

(x,y) = (-0.149,22.978) and (20.149,-382.982)

y' = 0.290 and -40.298

the line through (-0.149,22.978) and (10,26) has slope 3.022/10.149 = 0.29

the line through (20.149,-382.982) and (10,26) has slope 408.982/-10.149 = 40.298

Look slike those are our lines.