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Q1- Find the x-coordinate of the point on the graph of y=x^2 where the tangent line is parallel to the secant line that cuts the curve at x=-1 and x=2
Q2- Solve | x −3|^2 −4| x −3|=12 for x.
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Q1:
1. Start with the secant.
First point has the coordinates (-1,1) and the second one (2,4).
2. Write down the ecuation of the secant.
For example (y-1)/(4-1)=(x+1)/(2+1) same as y-x-2=0
3. Find the gradient of the secant
m = 1
4. The tangent you are looking for is parallel with the secant so they have the same gradient m = 1
5. You have now the gradient, and you can solve an equation to find the point you need.
y = x^2 gives you y' = 2x
2x = 1 (=m)
x=1/2
Answer x=1/2
Check my calculations anyway. I can be wrong, but this is the idea.
Q2: if you make a notation /x-3/ = m (> 0) you get an equation m^2-4m-12=0.
This equation has 2 solutions m=6 and m=-2.
-2 is not positive so m=6 and x-3 = 6 or x-3 = -6.
Solutions x = 9 or x = -3.
Again, check my calculations.
Last edited by mathsmypassion (2008-12-11 07:36:47)
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