Prove that there is no line through the point (1,2) that is tangent to the curve y=4-x^2

Question

1. the derivative or the slope to the function or the function for the tangent line is -2x, using the power rule....however 
2. I understand that first(1,2) and (x,4-x^2) are similar..i.e if the graph exist between those points....
3. the slope would be {[2-(4-x^2)]/x-2} the whole equation would equal -2x because the slope of those two points is equal to the the tangent line...
4. then when solving them after making them equal...i haven't solved it that far can you help me please solve this question 

Answer 1

One approach to this problem is the following:

 

Any line through the point (1,2)   will have the form   y = mx + 2-m   for some value of m.

 

Consider the intersection of this line with the curve   y = 4 - x²   .   Since both left hand sides are the same (y) , at

the intersection:    4 - x²  = mx  + 2 - m.     This is a quadratic equation for x.   The quadratic formula can be used to solve it.     After some rearrangement, the quadratic formula gives

 

  x =  -m/2  ± (1/2) sqrt( (m+2)² + 4)

 Since (m+2)² +4 is always positive, the are two solutions for any value of m.   This means two points of intersection, so there is no tangent.  (The argument of the square root would be zero for a tangent.)

 

The analysis above leaves out the possibility of a vertical line for which m is undefined.    However,  a vertical line through the point (1,2) intersects the curve 4 - x²  at the point (1,3).   The slope of 4 - x² is not undefined (vertical) at this point, so the intersection in this case is not tangential. 

Answer 2

Given y = 4-x²

 

Then y' = -2x   This is the slope of the tangent line.

 

Another expression for the slope of the tangent line in question is obtained from using the slope formula between a point (x,y) on the curve (x, 4-x²) and the point (1,2).  

 

m = (y₁-y₂)/(x₁-x₂)   Also the two expressions for the slope must be equal.

 

so -2X = [(4-x²) - 2]/(x - 1) = (2-x²)/(x-1)

 

Solve using cross multiplication

 

-2x(x-1) = 2 - x²

-2x² +2x = 2 -x²

 

rearranging yields

 

x² -2x  = -2

 

We can solve this by completing the square  

 

x² -2x + 1 = -2 + 1

 

(x-1)² = -1

 

x-1 = +- i

x = 1+- i

 

Since the only solutions for x are complex, no real point on the curve exists such that the tangent line passes through (1,2)