(a) find the slope of the tangent to the curve y=3+4x^2-2x^2 at the point where x=a (b) find the equation of the tangent lines at the points (1,5) and (2,3) (c)

Question

(a) find the slope of the tangent to the curve y=3+4x^2-2x^2 at the point where x=a

(b) find the equation of the tangent lines at the points (1,5) and (2,3)

(c) graph the curve and both tangents 

 

Answer 1

(a) the function can be simplified to y = 3 + 2x^2.

 

     If this is a typo, stop reading and repost

     the correct function with the proper exponents, as this is a little unusual.

 

 

 THe first derivative is y' = 4x

 

The slope of the tangent line at x=a is 4a.

 

(b)   at (1,5) the tangent line has slope 4*1 = 4

       y = M x + b

 

       5 = (4)(1) + b

      5 = 4 + b

     b = 1

 

    the equation for the first tangent line is y = 4x + 1

 

 (2,3) is not on the function. The slope at x=2  is M=2*4 = 8

 

   Again, y = M x + b

 

            3  = 8 (2) + b

            3 = 16 + b

 

             b = -13

 

  THe equation of the tangent line is y =8x - 13

 

(c) use graph paper or goto desmos.com and graph the function and the two lines.

 

Answer 2

Hi Julia.

 

I think the original problem should have the "2x²" really be 2x³.

This makes both points lie on the curve.  So this is the I will use to solve this problem.

 

The slope of the tangent line is the derivative at each of the points listed.

 

So the derivative of  f(x)= 3 + 4x² - 2x³  is

 

f'(x) = 8x - 6x²

 

Using  (1,5), f'(1) = 2   so the slope of the tangent line at (1,5) is 2.

Using (2,3), f'(2) = -8     so the slope of the tangent line at (2,3) is -8.

 

(a)  The slope at x=a is f'(a) = 8a - 6a²

 

(b)   Knowing the slope at (1,5) is 2, the equation of the tangent line is found by simplifying the point-slope form of the equation of a line.  In this case, (y -5) = 2(x - 1)

 

        The slope at (2,3) is -8, so the point-slope form of the linear equation is  (y-3) = -8(x-2)

 

If your teacher wants you to, you can simplify these down to slope-intercept form of y=mx+b.

 

Sorry - can't graph on this forum!  :-)