find the equation of the line tangent to y = -5x + 3x at (2, -14)

Question

 I need help with this calculus question.

Answer 1

OK, let's assume the equation is:

 

      y = -5x² + 3x

 

If it isn't, the technique is the same - just apply it to whatever the actual equation is.  The equation of a line is:

 

      y = mx +b

 

Where m is the slope of the line and b is the y-intercept (the point where the line crosses the y-axis).  So to find the equation of the tangent line, we need to find the right values of m and b.  The key here is to remember that derivative of a function is the slope of the tangent line at every point - that is, it is the "m" (slope) of our tangent line..

 

     dy/dx = d(-5x²)/dx + d(3x)/dx = -10x + 3

 

At the point x= 2, the slope is m = dy/dx = -10*2 + 3 = -17.  So the equation of our tangent line at x=2 is:

 

     y = -17x + b

 

To find b, plug in the given point on the curve, which is also on the tangent line (2,-14):

 

     y = -17x + 8

     -14 = -17*2 + b

     -14 = -34 + b

     20 = b

 

So the final equation of the tangent line is:

 

     y = -17x + 20

Answer 2

Do you mean -5x^2?