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

• thomasrules
In summary, two tangent lines that pass through the point (2,8) and are tangent to the curve y=x^3 can be found by solving the equation 2x^3 - 6x^2 + 8 = 0 for x, which gives the solutions x=2 and x=-1. Plugging these values into the equation y - 8 = m(x-2) yields the equations y=12x-16 and y=3x+2 as the two tangent lines.
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

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)

isn't that the same thing...

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

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.

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

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

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thomasrules said:
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

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.

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.

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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

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.

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

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

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

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

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.

Max Eilerson said:
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

## 1. How do I find the equations of two lines that pass through the point (2,8) and are tangent to the curve y = x^3?

To find the equations of two tangent lines, we will use the fact that the slope of a tangent line is equal to the derivative of the curve at that point. So, first we will find the derivative of y = x^3, which is y' = 3x^2. Then, we will substitute the x-coordinate of the given point (2,8) into the derivative to find the slope of the tangent line. Next, using the slope and the given point, we can use the point-slope form to find the equation of each tangent line.

## 2. Can I use any point other than (2,8) to find the equations of the tangent lines?

Yes, you can use any point on the curve y = x^3 to find the equations of the tangent lines. Just remember to use the corresponding x-coordinate in your calculations.

## 3. How many tangent lines can be drawn through the point (2,8) and tangent to y = x^3?

There can be two tangent lines drawn through the point (2,8) and tangent to y = x^3. This is because the curve y = x^3 is a cubic function, which can have two tangent lines at a single point.

## 4. Can you graphically show the two tangent lines through the point (2,8) and tangent to y = x^3?

Yes, you can graphically show the two tangent lines by plotting the point (2,8) on the curve y = x^3 and drawing two tangent lines passing through that point. The slope of the tangent lines will be equal to the derivative of the curve at that point, which is 12. You can use the slope-intercept form or the point-slope form to graph the lines.

## 5. How can I verify that the two tangent lines found are indeed tangent to the curve y = x^3?

To verify that the two tangent lines are indeed tangent to the curve y = x^3, you can substitute the x-coordinate of the point of tangency (2,8) into the original curve and the tangent lines. If the resulting y-values are the same, then the lines are tangent to the curve at that point. Additionally, you can also check if the slope of the tangent lines are equal to the derivative of the curve at that point, which is 12 in this case.

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