Find Equations Of Both Lines Through The Point (2, −3) That Are Tangent To The Parabola Y = X2 + X.

Find the equation of both lines that pass through the point (2, -3) and are tangent to the parabola y = x^2 + x.

We have to find the equations of both lines through the point (2,-3) that are tangent to the parabola y=x^2+x

First we will  find the derivative of y to find the slope m.

i.e. m=y'=2x+1

Now we find two points on the two tangent lines to y using: (x,x^2+x) and (2,-3). This is to show that (x,f(x)) is a point on the original equation f(x).

Now we have to find the slope of these two points and equate it to the derivative or slope of the original equation.

i.e. 2x+1=\frac{x^2+x+3}{x-2}

implies, x^2-4x-5=0
i.e. x=5 , -1 

So now at x=5 the slope is m= 2(5)+1=11.So, the equation of the first tangent line through point (2,-3) is y+3=11(x-2)
i.e. \mathbf{y=11x-25}

at x=-1, the slope is m=2(-1)+1=-1 .So, the equation of the second tangent line through point (2,-3) is: y+3= -1(x-2).
i.e.  \mathbf{y=-x-1}

So the equation of the two tangent lines to the parabola y=x^2+x passing through the point (2,-3) are:

y=11x-25 and y=-x-1

The slope of the line tangent...

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