SOLUTION: Find the equation of the tangent to the curve y=x+x^2 at the point where x=a find the values of a for which this line passes through the point P(2,-3).Hence find the equation of

Question 1149667: Find the equation of the tangent to the curve y=x+x^2 at the point where x=a find the values of a for which this line passes through the point P(2,-3).Hence find the equation of the tangent from P to the curve.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52847) (Show Source): You can put this solution on YOUR website!
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The derivative of the given function is y'(x) = 2x+1; its value at x= a is y'(a) = 2a+1. The line should go through the points (a,a^2+a) and P(2,-3). Therefore, the slope of this line is m = = So, the equation to find the value of "a" is 2a+1 = . saying that the two expressions for the slope are equal. Simplify and solve for "a". First step is to multiply both sides by (a-2) (2a+1)*(a-2) = a^2 + a + 3 2a^2 + a - 4a - 2 = a^2 + a + 3 2a^2 - 3a - 2 = a^2 + a + 3 a^2 -4a - 5 = 0. (a-5)*(a+1) = 0 The roots are a= 5 and a= -1. Hence, there are two points on the curve satisfying the given conditions - one point is (5,30) and the other point is (-1,0), and two such lines : - one with the slope of 2*5+1 = 11 and the other with the slope of 2(-1)+1 = -1. The first line has the equation y+3 = 11*(x-2), or y = 11x - 25. The second line has the equation y+3 = -(x-2), or y = -x -1. Plot y = x+x^2 (red), y = 11x-25 (green) and y = -x-1 (blue).

Solved.

Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

I do not understand the instructions as you show them for solving the problem, so I will ignore them....

The given function is y = x+x^2.

The derivative of the function is y' = 1+2x.

An arbitrary point on the curve is (a,a+a^2).

The derivative at the point (a,a+a^2) is 1+2a.

We want the slope of the curve at (a,a+a^2) to be equal to the slope between (2,-3) and (a,a+a^2).

or

The two points of tangency are where a=5 and where a=-1. Those points are (5,30) and (-1,0).

A graph, showing the points of tangency at x=-1 and x=5.

If you need the equations of the tangent lines, use the two known points (-1,0) and (2,-3) for one of them and the two known points (5,30) and (2,-3) for the other.

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