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| Question 1017822:  Sketch the curve y=x^3-9x^2+15x-7 indicationg clearly it point of intercetion with the axes and its turning points.
 THANK YOU
 Answer by Alan3354(69443)
      (Show Source): 
You can put this solution on YOUR website! Sketch the curve y=x^3-9x^2+15x-7 indicationg clearly it point of intercetion with the axes and its turning points. ------------
 Pick values for x and find y.
 eg
 x = 0, y = -7 --> (0,-7)
 x = 1, y = 0 --> (1,0)
 Plot as many points as you like, then draw a curve thru them.
 ----
 Use Excel to make a table if you have that.
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 The intersection with the y-axis is at x = 0, shown above (0,-7)
 One intersection with the x-axis is (1,0) shown above.
 x^3-9x^2+15x-7 = (x-1)*(x-1)*(x-7)
 --> intersections at x = 1, 1, 7
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 The 1st derivative y' = 3x^2 - 18x + 15
 --> x^2 - 6x + 5 = 0
 (x-1)*(x-5) = 0
 --> turning points at x = 1 and x = 5
 
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