SOLUTION: Find the gradient of the tangent to y(x) = x2 at A(1,1), B(-1,1) and C(2,4).

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Find the gradient of the tangent to y(x) = x2 at A(1,1), B(-1,1) and C(2,4).      Log On


   

Question 885173: Find the gradient of the tangent to y(x) = x2 at A(1,1), B(-1,1) and C(2,4).
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
WITH CALCULUS:
In calculus, you are told that the slope/gradient of the tangent at a certain point
is the value of the derivative of the function at that point.
The derivative of x%5En is n%2Ax%5E%28n-1%29 , so
the derivative of y%28x%29=x%5E2 is
dy%2Fdx=2x .
At point A(1,1), x=1 so dy%2Fdx=2%2A1=highlight%282%29 .
At point A(-1,1), x=-1 so dy%2Fdx=2%2A%28-1%29=highlight%28-2%29 .
At point A(2,4), x=2 so dy%2Fdx=2%2A2=highlight%284%29 .

WITHOUT CALCULUS:
If you are not studying calculus,
you could write the equation of a line passing through the point given with slope m ,
and then you would find the value of m that gives that line
exactly one intersection point with y%28x%29=x%5E2 .
You would do that for each point.

For point A(1,1), a point with x%5BA%5D=red%281%29 and y%5BA%5D=blue%281%29 ,
the line would be y-blue%281%29=m%28x-red%281%29%29
Then, knowing that the tangent intersects the graph of y%28x%29=x%5E2 at only one point,
you would know that there should be just one solution to x%5E2-blue%281%29=m%28x-red%281%29%29 .
x%5E2-1=m%28x-1%29-->x%5E2-1=mx-m-->x%5E2-1-mx%2Bm=0-->x%5E2-mx%2B%28m-1%29=0
has just one solution when
%28-m%29%5E2-4%2A1%2A%28m-1%29=0-->m%5E2-4m-4=0-->%28m-2%29%5E2=0-->m-2=0-->m=highlight%282%29 .

For point B(-1,1), a point with x%5BA%5D=red%28-1%29 and y%5BA%5D=blue%281%29 ,
the line would be y-blue%281%29=m%28x-red%28-1%29%29 .
The equation to find the intersection point of that line and y%28x%29=x%5E2 is
x%5E2-blue%281%29=m%28x-red%28-1%29%29-->x%5E2-1=m%28x%2B1%29-->x%5E2-1=mx%2Bm%29-->x%5E2-1-mx-m=0%29-->x%5E2-mx%2B%28-m-1%29=0%29 .
For that equation to have just one solution, it must be
%28-m%29%5E2-4%2A1%2A%28-m-1%29=0-->m%5E2%2B4m%2B4=0-->%28m%2B2%29%5E2=0m%2B2=0-->m=highlight%28-2%29 .

For point C(2,4), a point with x%5BA%5D=red%282%29 and y%5BA%5D=blue%284%29 ,
the line would be y-blue%284%29=m%28x-red%282%29%29 .
The equation to find the intersection point of that line and y%28x%29=x%5E2 is
x%5E2-blue%284%29=m%28x-red%282%29%29-->x%5E2-4=m%28x-2%29-->x%5E2-4=mx-2m%29-->x%5E2-4-mx%2B2m=0%29-->x%5E2-mx%2B%282m-4%29=0%29 .
For that equation to have just one solution, it must be
%28-m%29%5E2-4%2A1%2A%282m-4%29=0-->m%5E2-8m%2B16=0-->%28m-4%29%5E2=0m-4=0-->m=highlight%284%29 .