Question

I know how to find the equation of a tangent line given the equation of a line and a point on it but I'm not sure how to answer this question:

1) Find the equation of the tangent line to y=x^2 if the x-intercept of the tangent is 2.

or this one:

2) Show that the tangent to y=x^2 at a point (Xo,Yo) other than the vertex always has x-intercept (1/2)Xo.

I used the difference quotient to work out that the derivative / slope of tangent is 2x but I'm not sure where to go from there.

Answer 1

So in the first question, they give you the function they want you to find the tangent to. And they give you a point on the tangent line. But what they do not give you is the point *at* which the tangent line is actually tangent to the curve. If you are online, come back and we will work through this.

Answer 2

Ok so the slope of the tangent is 2(Xo) where Xo is some x-value on y=x^2 right? (I know it doesn't relate to the x in the tangent line equation) And I've been using point slope form for my equations. But even with a point (2,0) on the tangent line and the slope, 2(Xo) where Xo is an x-val on y=x^2, it seems like I'm still in need of a variable to be able to get the equation for the tangent line.

So I have the equation below with two variables and I'm not sure what to do.

Y-Yo = m(X-Xo) => Y-0 = 2Xo(X-2) => Y = 2(Xo)X - 4(Xo)

Answer 3

Yeah, it does seem under parametrized now that I look at it. I'm drawing it out...

Answer 4

I wrote out two points for the tangent (2,0) and (0, b) and used them write an equation for m. Specifically, m = (-1/2)b. You can rearrange that for b to get b = -2m and substitute that to parametrize the rest of the problem.

I got y = 8x - 16 for the equation for the tangent.

Answer 5

Thanks! That makes sense. Didn't think of using b.

Answer 6

You're quite welcome. I didn't either for quite some time. I am a biologist not a hard core math type... :D

Answer 7

Thanks for your patience.

Answer 8

Well given any line of slope \(m\) that must intersect the point \((x_0,y_0)\), the equation is given as follows.\[y-y_0=m(x-x_0)\]The \(x\)-intercept is given by substituting \(y=0\) and solving for \(x\).\[0-y_0=m(x-x_0) \Rightarrow \boxed{x=x_0-\dfrac{y_0}{m}}\ \text{($x$-intercept)}\]This should give you enough of a nudge to help you solve the problems.

Answer 9

slope of the tgnt=dy/dx=2x=2x_1 at (x_1,2x_1^2)
y=mx+c --> y=(2x1)x+(x1 ^2-2x1 ^2)=2(x_1)x-(x_1)^2