Question 1067725
{{{(dy/dx)(x^2-2)}}}
{{{m=2x}}}, slope for any line tangent to y=x^2-2.



{{{y+5=m(x-1)}}}, a line equation
{{{y=m(x-1)-5}}}
{{{y=(2x)(x-1)-5}}}
{{{y=2x^2-2x-5}}}, a line expected to be tangent to y=x^2-2.


Expression for the parabola, MINUS the expression for the tangent line, should be 0, where the line and parabola intersect.  This line is not above the parabola.


{{{x^2-2-(2x^2-2x-5)=0}}}, to solve for finding either of the intersection points.


{{{x^2-2-2x^2+2x+5=0}}}

{{{-x^2+2x+3=0}}}

{{{x^2-2x-3=0}}}

{{{(x+1)(x-3)=0}}}



If x=-1, then y=(-1)^2-2=-1.
Point on parabola, (-1,-1).


If x=3, then y=(3)^2-2=9-2=7.
Point on parabola, (3,7).



The Two Possible Lines Tangent to y=x^2-2, and containing (1,-5):
-
Line containing  (1,-5) and (-1,-1).
slope {{{(-1+5)/(-1-1)=4/(-2)=-2}}};
Equation starting point-slope form,
{{{highlight(y+5=-2(x-1))}}}


-
Line containing (1,-5) and (3,7).
slope {{{(12/2)=6}}};
Equation,  
{{{highlight(y+5=6(x-1))}}}