SOLUTION: 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

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.
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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.
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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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