Find the slope of a line that passes thorugh the origin and.

[Me_self]

A Linear function which passes through the origin is tangent to the function y= x^2(6-x). find the slope of the line. (hint. draw a sketch of the scenario and then realize that the line will intersect the curve at only two point ( the origin and tangent point)..

help.. i tried graphing it on my calc.. and i got 9x but its just a guess i need something accurate please help

thanks in advance

 

[fasteddie65]

If y = 9x is a tangent line, then it should intersect the curve at only one other point besides the origin.

x^2(6 - x) = 9x
6x^2 - x^3 = 9x
0 = x^3 - 6x^2 + 9x
0 = x(x^2 - 6x + 9) = x(x - 3)^2

This equation has a zero at x = 0 and a double zero at x = 3, which would be a tangent point.

 

[galactus]

eddie done a fine job, but if I may. When you're confronted with a tangent line problem and they do not tell you where it is tangent, but all they give you is another line it passes through, you can try it this way.

Using \(\displaystyle y-y_{1}=m(x-x_{1})\)

We know \(\displaystyle y=6x^{2}-x^{3}\)

the slope is the derivative, so:

\(\displaystyle m=12x-3x^{2}\)

Plug everything into the general form from above along with (0,0), the point it passes through.


.............................................\(\displaystyle \text{x \;\ x1}\)
\(\displaystyle \underbrace{6x^{2}-x^{3}}_{\text{y}}-\overbrace{0}^{\text{y1}}=\underbrace{(12x-3x^{2})}_{\text{m=slope}}(\overbrace{x-0}^{\swarrow\searrow})\)

\(\displaystyle 2x^{2}(x-3)\Rightarrow x=0, \;\ x=3(\text{multiplicity 2})\)

Now, we have the line equation because we have the much needed x value.

Using the slope intercept form, \(\displaystyle y=mx+b\). Since it crosses at the origin, b=0

And we have \(\displaystyle \fbox{y=9x}\)

Just as Big Ed produced.