Quick tangent line question (calc final tomorrow)

In summary, the problem is to find all the tangent lines that pass through the point (0,0). To find the slope of the tangent line, the problem asks how many lines can be found in one point and then calculates the slope for each line.
  • #1
erjkism
54
0
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)
[tex]y=x^{3}+ 6x^{2}+8[/tex]i took the derivative and got this.
[tex]y=3x^{2}+12x[/tex]

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?
 
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  • #2
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.
 
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  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 
  • #7
Equation of a line: [tex] y_1-y_0 = m(x_1 - x_0) [/tex]
Two points on the line: [tex] (0, 0) (x,y) [/tex]
slope: [tex] m = y' [/tex]
original equation: [tex] y = x^3+6x^2 +8 [/tex]

Start filling in the blanks.
 
  • #8
erjkism said:
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)
[tex]y=x^{3}+ 6x^{2}+8[/tex]


i took the derivative and got this.
[tex]y=3x^{2}+12x[/tex]
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?

erjkism said:
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 [itex]y= mx= x^{3}+ 6x^{2}+8[/itex] and [itex]y'= m= 3x^2+ 12x[/itex]. 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.
 

What is a tangent line?

A tangent line is a line that touches a curve at only one point. It is perpendicular to the curve at that point and represents the instantaneous rate of change of the curve at that point.

Why is finding tangent lines important in calculus?

Finding tangent lines is important because it allows us to approximate the behavior of a curve at a specific point. This is useful in many applications, such as optimization problems and determining the slope of a curve at a certain point.

How do you find the equation of a tangent line?

To find the equation of a tangent line, you first need to find the slope of the curve at the point of tangency. This can be done by taking the derivative of the function at that point. Then, using the point-slope form of a line, you can plug in the point of tangency and the slope to find the equation of the tangent line.

What is the difference between a tangent line and a secant line?

A tangent line touches a curve at only one point, while a secant line intersects the curve at two points. The slope of a tangent line represents the instantaneous rate of change, while the slope of a secant line represents the average rate of change between the two points of intersection.

Can you have a horizontal tangent line?

Yes, a horizontal tangent line occurs when the slope of the curve at a certain point is equal to 0. This means that the curve is not changing at that point, and the tangent line is parallel to the x-axis.

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