Question number 5 - y=6x-12 and y=x^3-x^2-2x...

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The tangent line to the graph of y = x^3 - x^2 - 2x at x=2 intersects the graph of y = x^3 - x^2 - 2x at one another point. Find the intersection point. There are two steps you need to understand to figure out this problem: 1. Find the tangent line of y(x) at (2,y(2)). 2. Find what x equals to when you equate the tangent line and y(x). 1. To get the tangent line, find M (slope) by deriving y(x) and plugging in x=2 to get the slope of the tangent line. dy/dx = 3x^2-2x-2 dy/dx = 3(2)^2-2(2)-2 = 6 Use the line formula: y-y0 = m(x-x0) y-y(2)=6(x-2) y-0=6(x-2) y=6(x-2) Now equate your tangent line to the original equation:
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Unformatted text preview: y=6x-12 and y=x^3-x^2-2x 6(x-2)=x^3-x^2-2x Factor the right side to find where x=0, 6(x-2)=x(x-2)(x+1) We can divide both sides by (x-2) 6=x(x+1) Multiply the right side out and equate the formula to 0 to find where x=0 6=x^2+x 0=x^2+x-6 0=(x-2)(x+3) So 0=x-2 => x=2 tells us that y will intersect at x=2, which was the original point, and 0=x+3 => x=-3 tells us that y will also intersect at x=-3. To get the intersection point, just plug x=-3 back into your original equation to get y: y(-3)=(-3)^3-(-3)^2-2(-3) y(-3)=-27-9+6 y(-3)=-30 So your point is (-3, -30)....
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This note was uploaded on 03/24/2011 for the course MATHEMATIC 142 taught by Professor Prof.bruceabgarian during the Spring '11 term at Nassau CC.

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