# Finding tangent to function parallell with line

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• Dec 1st 2009, 02:16 PM
Bato91
Finding tangent to function parallell with line
Hello all! I just found this forum and registered. I would like some help with the following problem:

What is the function of the tangent to the parable $\displaystyle y=x^2$ , parallell with the line $\displaystyle y=6x$ ? I can see where the tangent is graphically and go from there, but how can you solve it solely by equating?

(feel free not to copy my terminology; I'm not good at English and may use some odd names)
• Dec 1st 2009, 03:09 PM
qmech
Derivative
Your function
$\displaystyle y = 6 x$

has slope 6.

Your function
$\displaystyle y = x ^ 2$

has derivative:
$\displaystyle \frac { d y } { dx } = 2 x$

which has slope 6 at x=3. Thus y = 9. The tangent line must be y=6x-9.
• Dec 2nd 2009, 12:14 AM
Bato91
Thank you!
• Dec 2nd 2009, 12:50 AM
earboth
Quote:

Originally Posted by Bato91
Hello all! I just found this forum and registered. I would like some help with the following problem:

What is the function of the tangent to the parable $\displaystyle y=x^2$ , parallell with the line $\displaystyle y=6x$ ? I can see where the tangent is graphically and go from there, but how can you solve it solely by equating?

(feel free not to copy my terminology; I'm not good at English and may use some odd names)

Here is a different approach:

The tangent line has the equation: $\displaystyle t: y = 6x + b$

Now calculate the coordinates of the intersection between the parabola and the straight line:

$\displaystyle x^2=6x+b~\implies~x^2-6x-b=0$

Use the formula to solve this quadratic equation:

$\displaystyle x = 3\pm\sqrt{9+b}$

You get only one intersection point (=tangent point) if the radicand equals zero:

$\displaystyle 9+b = 0~\implies~\boxed{b=-9}$ . Consequently the tangentpoint is T(3, 9)

Therefore the equation of the tangent line becomes: $\displaystyle t: y = 6x-9$