Find two lines through the point (2,8) and tangent to y = x^3

[thomasrules]

Find two lines that pass through the point$$(2,8)$$ and are tangent to the curve $$y=x^3$$

I found the first equation which is $$y=12x-16$$ by finding the derivative of $$y=x^3$$ which is $$y=3x^2$$.

I then plugged in $$(2,8)$$ into $$y=mx+b$$ along with the slope of $$12$$.

But how can there be more than one slope at a single point? I don't get it

 

[arildno]

Does the text say that the lines has to be tangent to the given curve AT the point (2,8)?
From how I read it, the text just requires that the lines go THROUGH (2,8)

 

[thomasrules]

isn't that the same thing...

and there only should be 1 tangent for an equation where the answer there's 2

 

[z-component]

arildno is saying that the problem not not be asking for two tangent lines through the same point, but instead a tangential line and a non-tangential line to (2,8). If that's the case, this is simple to do. You already have the tangent's equation, so just make an equation for a different line that passes through (2,8) with a different slope.

 

[alfredbester]

$$y - 8 = m(x-2)$$
$$y = (3x^2)(x-2) + 8$$
$$x^3 = (3x^2)(x-2) + 8$$
$$2x^3 - 6x^2 + 8 = 0$$

Solve for x and then use this to find equation of the two lines.

 

[thomasrules]

Find two lines that pass through the point$$(2,8)$$ and are tangent to the curve $$y=x^3$$

so both of them have to be tangent...yea i guess that makes sense what u said but it seems that is not what the question is asking

 

[radou]

The second line has to be a tangent too, and it has to pass through the point (2, 8). So, these are two conditions.

Use the equation of a line which passes through two points (x1, y1) and (x2, y2): y - y1 = ( (y2 - y1)/(x2 - x1) ) (x - x1). So, for example, set (x1, y1) = (2, 8). Further on, y2 must equal x2^3, and x2 is the point you need to find.

Further on, the equation of a tangent line at a point c is:
y - f(c) = f'(c) (x - c). So, set c = x2.

Now bring both of the equations into the form y = m x + b. Set the two slopes equal, and you'll get your point x2, which you can plug back into any of the equations to obtain the other tangent.

 

[Max Eilerson]

The method posted by alfredbester should work fine no? Solve the cubic, put the x-values back into equation for y. Then use (x2. y2) and (x3, y3) and y = mx + c.

 

[thomasrules]

ok answers are: y=12x-16 and y=3x+2 that means that there are 2 slopes for the tangent at (2,8) which is impossible

 

[Max Eilerson]

This isn't true at all. Draw the x^3 graph then try to draw straight lines that pass through a specific point, can you draw two lines that got through the point and then make a tangent to the curve? The line does not need to be tangent to the curve at the point (2,8) just pass the point (2,8) and later be tangent with the curve.

 

[thomasrules]

so I got x=2 or a=2 but I'm stuck from here
so

$$x^3=3x^2(2)+b\ \ \ \ \ \ \ \ or\ \ a^3=3a^2(2)+b\ \ \ \ \ \ \ \ \ 8=24+b, b= -16$$

$$y=12x-16$$ and I already got that...how do I get the other

 

[Max Eilerson]

Stuck on what? You already have the equation of the two lines tangent to the curve that pass through (2,8).

 

[thomasrules]

so I got x=2 or a=2 but I'm stuck from here
so

$$x^3=3x^2(2)+b\ \ \ \ \ \ \ \ or\ \ a^3=3a^2(2)+b\ \ \ \ \ \ \ \ \ 8=24+b, b= -16$$

$$y=12x-16$$ and I already got that...how do I get the other

 

[Max Eilerson]

$$y - 8 = m(x-2)$$
$$y = (3x^2)(x-2) + 8$$
$$x^3 = (3x^2)(x-2) + 8$$
$$2x^3 - 6x^2 + 8 = 0$$

Solve for x and then use this to find equation of the two lines.

So you have that, solve for x. You will get X= 2, -1.
Then use y = mx + c, to find the two equations which you've already written down.

 

[thomasrules]

So you have that, solve for x. You will get X= 2, -1.
Then use y = mx + c, to find the two equations which you've already written down.

holy christ I'm dumb, I only took the roots of (x-2)(x-2)

thank you max