Point A lies on the curve y = (2x +1)^3 + 1. If the y coordinate of A is 2, find the equations of the tangent and normal to the curve at A. If the tangent and normal at A meets the x-axis at B and C respectively, find the area of the triangle ABC.

Question

anonymous · Answer

ok.  step 1, get the value of x at point A.  $$2=(2x+1)^3+1$$Solving for x:$$x=0$$Next, find the derivative:$$y'=6(2x+1)$$This will be the slope of your tangent line to the graph.  Plugging in x=0 you get y'=6.  The equation of a line through the point (0,2) with slope y'=6 is$$(y-2)=6(x-0)$$Or,$$y=6x+2$$

anonymous · Answer

The line normal to the curve at the point (0,2) will have a slope of $$m=\frac{-1}{6}$$ the perpendicular line will have a slope that is the negative reciprocal of the tangent lines slope.  Use this to get the equation of the normal line.

anonymous · Answer

Now what is point B?  Well, if this is where the tangent line meets the x axis, it is the x-intercept of the line y=6x+2.  So,it must be the point (-1/3, 0)

anonymous · Answer

You should e able to get the rest from here :)

anonymous · Answer

I got these answers too but was unsure as the answers are in fractions. My area of triangle ABC is 12 1/3 or 12.3333. Was it right?

anonymous · Answer

The normal line equation should be $$y=\frac{-1}{6}x+2$$So point C is at (12,0). Vector BC is <37/3,0> and vector BA is <-37/3,2>. Area of the triangle ABC is 37/3 or 12 1/3. Looks correct :)