Question 885173
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^n}}} is {{{n*x^(n-1)}}} , so
the derivative of {{{y(x)=x^2}}} is
{{{dy/dx=2x}}} .
At point A(1,1), {{{x=1}}} so {{{dy/dx=2*1=highlight(2)}}} .
At point A(-1,1), {{{x=-1}}} so {{{dy/dx=2*(-1)=highlight(-2)}}} .
At point A(2,4), {{{x=2}}} so {{{dy/dx=2*2=highlight(4)}}} .
 
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(x)=x^2}}} .
You would do that for each point.
 
For point A(1,1), a point with {{{x[A]=red(1)}}} and {{{y[A]=blue(1)}}} ,
the line would be {{{y-blue(1)=m(x-red(1))}}} 
Then, knowing that the tangent intersects the graph of {{{y(x)=x^2}}} at only one point,
you would know that there should be just one solution to {{{x^2-blue(1)=m(x-red(1))}}} .
{{{x^2-1=m(x-1)}}}-->{{{x^2-1=mx-m}}}-->{{{x^2-1-mx+m=0}}}-->{{{x^2-mx+(m-1)=0}}}
has just one solution when
{{{(-m)^2-4*1*(m-1)=0}}}-->{{{m^2-4m-4=0}}}-->{{{(m-2)^2=0}}}-->{{{m-2=0}}}-->{{{m=highlight(2)}}} .
 
For point B(-1,1), a point with {{{x[A]=red(-1)}}} and {{{y[A]=blue(1)}}} ,
the line would be {{{y-blue(1)=m(x-red(-1))}}} .
The equation to find the intersection point of that line and {{{y(x)=x^2}}} is
{{{x^2-blue(1)=m(x-red(-1))}}}-->{{{x^2-1=m(x+1)}}}-->{{{x^2-1=mx+m)}}}-->{{{x^2-1-mx-m=0)}}}-->{{{x^2-mx+(-m-1)=0)}}} .
For that equation to have just one solution, it must be
{{{(-m)^2-4*1*(-m-1)=0}}}-->{{{m^2+4m+4=0}}}-->{{{(m+2)^2=0}}}{{{m+2=0}}}-->{{{m=highlight(-2)}}} .
 
For point C(2,4), a point with {{{x[A]=red(2)}}} and {{{y[A]=blue(4)}}} ,
the line would be {{{y-blue(4)=m(x-red(2))}}} .
The equation to find the intersection point of that line and {{{y(x)=x^2}}} is
{{{x^2-blue(4)=m(x-red(2))}}}-->{{{x^2-4=m(x-2)}}}-->{{{x^2-4=mx-2m)}}}-->{{{x^2-4-mx+2m=0)}}}-->{{{x^2-mx+(2m-4)=0)}}} .
For that equation to have just one solution, it must be
{{{(-m)^2-4*1*(2m-4)=0}}}-->{{{m^2-8m+16=0}}}-->{{{(m-4)^2=0}}}{{{m-4=0}}}-->{{{m=highlight(4)}}} .