SOLUTION: The question asks:
The tangent line intersects the graph of f(x)=2x^3+10x^2−28x at another point. Determine the other point of intersection. I have already found my tangent
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Question 1090061: The question asks:
The tangent line intersects the graph of f(x)=2x^3+10x^2−28x at another point. Determine the other point of intersection. I have already found my tangent line eq'n y=-34x+18 @ ×= -3. I substituted and factored :
-34x+18=2x^3+10x^2-28x=0
2x^3+10x^2+6x-18=0
2(x^3+5x^2+3x-9)=0
2(x-1)(x+3)^2=0
I know that my x-axis is 1, how do I find y-coordinate? Or do I need to ?
Answer by natolino_2017(77) (Show Source): You can put this solution on YOUR website!
You can evaluate on f or in the tangent line.
y(1) = f(1) = -34 + 18 = 2 + 10 -28 = 16.
So the second point is (1,16) although it has a different slope on that point (f'(1) = 6+20-28= -2 which is not -34)
@natolino_
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