Quick tangent line question (calc final tomorrow)

[erjkism]

1. The problem statement, all variables and given/known
i have to find all tangent lines in the equation below that pass though the point (0,0)
$$y=x^{3}+ 6x^{2}+8$$i took the derivative and got this.
$$y=3x^{2}+12x$$

then i substituted the points x=0 and y=0 into the derivative equation and the got 0=0.

i am kind of stuck here cause i haven't done a problem like this in a while. what should i do next?

 

[Dick]

The derivative equation is y'=3x^2+12x, that's the slope of the tangent line. Another way to express the slope of the tangent line is
(delta y)/(delta x)=(y-0)/(x-0)=(x^3+6x^2+8)/x. Equate the two.

 

[erjkism]

i know that the derivative is the slope/ but how can i find the equations of all of the tangent lines that go thru the origin?

 

[Dick]

There are two different way to express the slope of the tangent line. i) use the derivative, ii) use the difference (delta y)/(delta x) for two points on the line, like (x,x^3+6x^2+8) and (0,0). Isn't that what I just said? Equate them.

 

[ace123]

Well since you found the slope and you have the point (0,0) why not just do the point slope forumla? That will give you the equation of the tangent line. Right?

 

[Petkovsky]

First of all ask yourself how many tangent lines can you find in one point.
Second, you have found the formula for the slope of the curve in ANY point, but you need to find the one for X=0, and then construct a line that has that slope and passes through the point that has been given to you.

 

[sennyk]

Equation of a line: $$y_1-y_0 = m(x_1 - x_0)$$
Two points on the line: $$(0, 0) (x,y)$$
slope: $$m = y'$$
original equation: $$y = x^3+6x^2 +8$$

Start filling in the blanks.

 

[HallsofIvy]

1. The problem statement, all variables and given/known
i have to find all tangent lines in the equation below that pass though the point (0,0)
$$y=x^{3}+ 6x^{2}+8$$


i took the derivative and got this.
$$y=3x^{2}+12x$$

No, that's y', not y.

then i substituted the points x=0 and y=0 into the derivative equation and the got 0=0.

The problem does not say the curve passes through (0,0) (nor is it true). There is no point in putting x= 0, y= 0 in any equation.

i am kind of stuck here cause i haven't done a problem like this in a while. what should i do next?

i know that the derivative is the slope/ but how can i find the equations of all of the tangent lines that go thru the origin?

Any line that goes through (0,0) is of the form y= mx and has slope m. The tangent line must go through a point on the graph and must have m equal to the derivative. At that point $y= mx= x^{3}+ 6x^{2}+8$ and $y'= m= 3x^2+ 12x$. Solve those two equation for m and x (it's only m you need but different values of x may give different slopes and different tangent lines). Try multiplying both sides of the second equation by x and setting the two values for mx equal.