tangent lines: for f(x) = kx^2+c, find x0 in terms of k so

[Lokito]

Let f(x) = kx[sup:38n5zlal]2[/sup:38n5zlal] + c

(a) Find x[sub:38n5zlal]0[/sub:38n5zlal] in terms of k such that the tangent lines to the graph of f at (x[sub:38n5zlal]0[/sub:38n5zlal], f(x[sub:38n5zlal]0[/sub:38n5zlal])) and (-x[sub:38n5zlal]0[/sub:38n5zlal], f(-x[sub:38n5zlal]0[/sub:38n5zlal])) are perpendicular.

(b) Find the slopes of the tangent lines mentioned in (a).

(c) Find the coordinates, in terms of k and c, of the point of intersection of the tangent lines mentioned in (a).

What is this question asking in part (a)? I don't know what x[sub:38n5zlal]0[/sub:38n5zlal] is referring to.

 

[Deleted member 4993]

What is this question asking in part (a)? I don't know what x[sub:31so8af4]0[/sub:31so8af4] is referring to.

x[sub:31so8af4]0[/sub:31so8af4] is just a point that you need to find.

The slope of the tangent through [x[sub:31so8af4]0[/sub:31so8af4],f(x[sub:31so8af4]0[/sub:31so8af4])] is = f'(x[sub:31so8af4]0[/sub:31so8af4]) = m[sub:31so8af4]1[/sub:31so8af4] = 2*k*x[sub:31so8af4]0[/sub:31so8af4]........... ..........(1)

The slope of the tangent through [-x[sub:31so8af4]0[/sub:31so8af4],f(-x[sub:31so8af4]0[/sub:31so8af4])] is = f'(-x[sub:31so8af4]0[/sub:31so8af4]) = m[sub:31so8af4]2[/sub:31so8af4] = -2*k*x[sub:31so8af4]0[/sub:31so8af4]........... ..........(2)

These tangent lines are perpendicular to each other - so what is the relationship between m[sub:31so8af4]1[/sub:31so8af4] & m[sub:31so8af4]2[/sub:31so8af4].

 

[Lokito]

What is this question asking in part (a)? I don't know what x[sub:kfh8b0xo]0[/sub:kfh8b0xo] is referring to.

x[sub:kfh8b0xo]0[/sub:kfh8b0xo] is just a point that you need to find.

The slope of the tangent through [x[sub:kfh8b0xo]0[/sub:kfh8b0xo],f(x[sub:kfh8b0xo]0[/sub:kfh8b0xo])] is = f'(x[sub:kfh8b0xo]0[/sub:kfh8b0xo]) = m[sub:kfh8b0xo]1[/sub:kfh8b0xo] = 2*k*x[sub:kfh8b0xo]0[/sub:kfh8b0xo]........... ..........(1)

The slope of the tangent through [-x[sub:kfh8b0xo]0[/sub:kfh8b0xo],f(-x[sub:kfh8b0xo]0[/sub:kfh8b0xo])] is = f'(-x[sub:kfh8b0xo]0[/sub:kfh8b0xo]) = m[sub:kfh8b0xo]2[/sub:kfh8b0xo] = -2*k*x[sub:kfh8b0xo]0[/sub:kfh8b0xo]........... ..........(2)

These tangent lines are perpendicular to each other - so what is the relationship between m[sub:kfh8b0xo]1[/sub:kfh8b0xo] & m[sub:kfh8b0xo]2[/sub:kfh8b0xo].

So I just put the equations into point slope and set them equal to each other?

y - (kx[sub:kfh8b0xo]0[/sub:kfh8b0xo][sup:kfh8b0xo]2[/sup:kfh8b0xo] + c) = 2kx[sub:kfh8b0xo]0[/sub:kfh8b0xo](x - x[sub:kfh8b0xo]0[/sub:kfh8b0xo])
and
y - (-kx[sub:kfh8b0xo]0[/sub:kfh8b0xo][sup:kfh8b0xo]2[/sup:kfh8b0xo] + c) = -2kx[sub:kfh8b0xo]0[/sub:kfh8b0xo](x + x[sub:kfh8b0xo]0[/sub:kfh8b0xo])

then solve for k?

I would then find the coordinates for part (c) by solving the same equations for y?

I think I'm still not getting it.

 

[Deleted member 4993]

Re: tangent lines

It will be much easier - if you answer the following part first:

These tangent lines are perpendicular to each other - so what is the relationship between m1 & m2?

 

[Lokito]

Re: tangent lines

It will be much easier - if you answer the following part first:

These tangent lines are perpendicular to each other - so what is the relationship between m1 & m2?

negative reciprocals - but that doesn't help me find x[sub:1lovcqh8]o[/sub:1lovcqh8] in terms of k.

I can't solve using the slopes - I'm going to get 2kx[sub:1lovcqh8]0[/sub:1lovcqh8] = 2kx[sub:1lovcqh8]0[/sub:1lovcqh8] , x[sub:1lovcqh8]0[/sub:1lovcqh8] = x[sub:1lovcqh8]0[/sub:1lovcqh8]

I'm looking for an x value where the tangent lines are perpendicular, meaning they have negative reciprocal slopes. I know that, but I don't see how to get there.

 

[stapel]

I can't solve using the slopes - I'm going to get 2kx[sub:1yhfxbqq]0[/sub:1yhfxbqq] = 2kx[sub:1yhfxbqq]0[/sub:1yhfxbqq] , x[sub:1yhfxbqq]0[/sub:1yhfxbqq] = x[sub:1yhfxbqq]0[/sub:1yhfxbqq]

Please reply with your demonstration of this.

You have found the formula for f'(x), you have plugged x[sub:1yhfxbqq]0[/sub:1yhfxbqq] into this, and found the expression for slope m[sub:1yhfxbqq]1[/sub:1yhfxbqq]. You have then plugged -x[sub:1yhfxbqq]0[/sub:1yhfxbqq] into f'(x), and found the expression for slope m[sub:1yhfxbqq]2[/sub:1yhfxbqq]. You have found the negative reciprocal for one of these slopes, and set it equal to the other slope. And... then what?

Please be complete. Thank you!

Eliz.

 

[Lokito]

Re:

I can't solve using the slopes - I'm going to get 2kx[sub:107z6q8q]0[/sub:107z6q8q] = 2kx[sub:107z6q8q]0[/sub:107z6q8q] , x[sub:107z6q8q]0[/sub:107z6q8q] = x[sub:107z6q8q]0[/sub:107z6q8q]

Please reply with your demonstration of this.

You have found the formula for f'(x), you have plugged x[sub:107z6q8q]0[/sub:107z6q8q] into this, and found the expression for slope m[sub:107z6q8q]1[/sub:107z6q8q]. You have then plugged -x[sub:107z6q8q]0[/sub:107z6q8q] into f'(x), and found the expression for slope m[sub:107z6q8q]2[/sub:107z6q8q]. You have found the negative reciprocal for one of these slopes, and set it equal to the other slope. And... then what?

Please be complete. Thank you!

Eliz.

I made a mistake. However, it's still not coming out right.
2kx[sub:107z6q8q]0[/sub:107z6q8q](x-x[sub:107z6q8q]0[/sub:107z6q8q]) = -1 / (-2kx[sub:107z6q8q]0[/sub:107z6q8q](x + x[sub:107z6q8q]0[/sub:107z6q8q])
There's nowhere to go from here.
2kx[sub:107z6q8q]0[/sub:107z6q8q]x - 2kx[sub:107z6q8q]0[/sub:107z6q8q][sup:107z6q8q]2[/sup:107z6q8q] = 1 / (2kx[sub:107z6q8q]0[/sub:107z6q8q]x + kx[sub:107z6q8q]0[/sub:107z6q8q][sup:107z6q8q]2[/sup:107z6q8q]/2)

 

[Lokito]

2kx[sub:3e31y0pm]0[/sub:3e31y0pm] = -1 / -2kx[sub:3e31y0pm]0[/sub:3e31y0pm]

2kx[sub:3e31y0pm]0[/sub:3e31y0pm] = 1 / 2kx[sub:3e31y0pm]0[/sub:3e31y0pm]

x[sub:3e31y0pm]0[/sub:3e31y0pm] = (1 / 2kx[sub:3e31y0pm]0[/sub:3e31y0pm]) * (1 / 2k) = 1 / 4k[sup:3e31y0pm]2[/sup:3e31y0pm]x[sub:3e31y0pm]0[/sub:3e31y0pm]

?

 

[Deleted member 4993]

\(\displaystyle x_0^2 \, = \, \frac{1}{4k^2}\)

\(\displaystyle x_0 \, = \,\pm \frac{1}{2k}\)

You really couldn't do this !!!

 

[Lokito]

\(\displaystyle x_0^2 \, = \, \frac{1}{4k^2}\)

\(\displaystyle x_0 \, = \,\pm \frac{1}{2k}\)

You really couldn't do this !!!

My ? was an "am I doing this right" kind of ?, not a "gar, algebra" one.

For parts (b) and (c), I need to find the derivative of the tangent lines (I think). I don't get this, though, because when you evaluate the original function's first derivative at a point to make the tangent lines you turn everything into constants. When x is x[sub:35buk65h]0[/sub:35buk65h], it's a constant. WIthout any variables in the equation I'm going to get a slope of 0 for both.

Part (c) would be setting the tangent lines equal to each other and solving.

 

[Deleted member 4993]

\(\displaystyle x_0^2 \, = \, \frac{1}{4k^2}\)

\(\displaystyle x_0 \, = \,\pm \frac{1}{2k}\)

You really couldn't do this !!!

My ? was an "am I doing this right" kind of ?, not a "gar, algebra" one.

For parts (b) and (c), I need to find the derivative of the tangent lines (I think)......Why - you know slopes [m[sub:3ktx2p77]1[/sub:3ktx2p77] & m[sub:3ktx2p77]2[/sub:3ktx2p77]] and you know the points the lines going through[{x[sub:3ktx2p77]0[/sub:3ktx2p77],f(x[sub:3ktx2p77]0[/sub:3ktx2p77])} & {-x[sub:3ktx2p77]0[/sub:3ktx2p77],f(-x[sub:3ktx2p77]0[/sub:3ktx2p77])} - now just find the equations of the lines

I don't get this, though, because when you evaluate the original function's first derivative at a point to make the tangent lines you turn everything into constants. When x is x[sub:3ktx2p77]0[/sub:3ktx2p77], it's a constant. WIthout any variables in the equation I'm going to get a slope of 0 for both.

Part (c) would be setting the tangent lines equal to each other and solving.