from the beginning of the coordinate system Going out a Tangent line to the Parabola:Y=9+(X-4)^2

find the Equation of the Tangent line

thnaks

Any particular point where it is tangent, or in general?

First, find the derivative of y to find the slope at some point.

Let's call the point of tangency (p,q).

y'=2x-8

Therefore, the slope at point (p,q) is m=2p-8.

Now, can you continue?

Your tangent goes through the point (0,0). Its equation therefore: y(x)=k*x, here k is its slope. This tangent (or these tangents - we don't know for sure how many) does two things. First, it has common point with the parabola. Second, both parabola and the tangent have the same slope at this point or, in other words, their derivatives are equal at this particular point. These two facts give you two equations.

k*x = 9 + (x-4)^2 (1)
k = 2*x - 8 (2)

Both equations have to hold for given k and x -- that is we have a system of two equations. Plug k from (2) into (1):

2*x^2 - 8*x = 9 + (x-4)^2

Solve this for x. How many solutions do you get? Each x is a point where the tangent starting from (0,0) may have common point with the parabola. From (2) you get k for your tangent equation or two.

Yes thanks to both of u I solved it.