If f (x) =-x^2+2, find the equation tangent to f (x) at the point (1,1)

Question

High school student 

Answer 1

To find the equation that is tangent, we must take the derivative of f(x).  The derivative is the slope of the tangent line.

 

f'(x) = -2x

 

Plug in the x value of the given coordinate.  x = 1.

 

f'(1) = -2(1)

       = -2

 

This is the slope of the tangent line.  The equation of the tangent line so far is

 

f(x) = -2x + b

 

 

Next, we plug in the coordinate into this tangent line equation to find b.

 

1 = -2(1) + b

 

1 = -2 + b

 

3 = b

 

 

The equation of the tangent line at this point is

 

f(x) = -2x + 3

Answer 2

Take the derivative of f(x) = -x^2 + 2

 

f'(x) = -2x

 

f'(1) = -2*1 = -2

 

The slope of the tangent linesoft the function at this point must be -2.

 

y = -2x + b

 

1 = -2*1 + b

 

1 = -2 + b

 

b = 3

 

y = -2x + 3 would be the tangent line to the function f(x) = -x^2 + 2 at the point (1,1).