First find the derivative of y.
derivative of y is 2x+1
and notice that derivative is slope of y.
Then find two points on the two tangent lines to y using:
(x,x2+x+4) and (2,1). This is to show (x,f(x)) is a point on the original equation f(x). Find the slope of these two points and equate it to the derivative (slope) of the original equation.
therefore 2x+1= (x2+x+4)-1 / x-2 to get x=5 or x=-1.
at x=5 the slope is 2(5)+1=11 and the point is (5,34) and the equation of the first tangent line is y-34=11(x-5)
at x=-1, the slope is 2(-1)+1=-1 and the point is (-1,4) and the equation of the second tangent line is
y-4= -1(x+1). It shows that every quadratic equation has two tangent lines from a point that is not on the graph of the quadratic equation.