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

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) (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 is , so
the derivative of is
.
At point A(1,1), so .
At point A(-1,1), so .
At point A(2,4), so .

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

For point A(1,1), a point with and ,
the line would be
Then, knowing that the tangent intersects the graph of at only one point,
you would know that there should be just one solution to .
-->-->-->
has just one solution when
-->-->-->--> .

For point B(-1,1), a point with and ,
the line would be .
The equation to find the intersection point of that line and is
-->-->-->--> .
For that equation to have just one solution, it must be
-->-->--> .

For point C(2,4), a point with and ,
the line would be .
The equation to find the intersection point of that line and is
-->-->-->--> .
For that equation to have just one solution, it must be
-->-->--> .
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