differentaition word problem: tangents to y=4x^2

[Becky4paws]

Find all the points (x,y) on the graph of the function y=4x^2 with the property that the tangent to the graph at (x,y) passes thru the point (2,0).

I went about this by first finding the derivative 8x. Wouldn't that be slope?

Then I would apply point/slope formula y-0=8x(x-2)?

I think I missed something.

 

[tkhunny]

You have made what I believe to be the most common error on this type of problem. You have assumed the given point is ON the curve. It isn't.

You have:
1) A point
2) A general expression for slope, referencing the first derivative.

You do NOT have:
1) The point(s) where your line(s) are tangent to the curve.

Where does that leave you?

 

[Gene]

Hint: Remember that y=4x² in your last equation.

 

[soroban]

Re: differentaition word problem

Hello, Becky4paws!

From the way you went way off in your first step,
\(\displaystyle \;\;\)I assume you could use a walk-through . . .

Find all the points \(\displaystyle (x,y)\) on the graph of the function \(\displaystyle y\,=\,4x^2\)
with the property that the tangent to the graph at \(\displaystyle (x,y)\) passes thru the point \(\displaystyle (2,0).\)

I don't suppose you made a sketch . . . [Naw, whatever for ??]

        *         |         */
                  |         /
                  |        o
           *      |      */
             *    |    * /
      -----------***----*------
                  |    /(2,0)
                  |   /

Here is one approach to this problem . . .

The line through \(\displaystyle (2,0)\) has slope \(\displaystyle m.\)
Its equation is: \(\displaystyle \,y\,-\,0\:=\:m(x\,-\,2)\;\;\Rightarrow\;\;y\:=\:mx\,-\,2m\)

Find the intersections of the parabola and this line:
\(\displaystyle \;\;4x^2\:=\:mx\,-\.2m\;\;\Rightarrow\;\;4x^2\,-\,mx\,+\,2m\:=\:0\)

Quadratic Formula: \(\displaystyle \,x\:=\:\frac{-(-m)\,\pm\,\sqrt{(-m)^2\,-\,4(4)(2m)}}{2(4)}\;= \;\frac{m\,\pm\,\sqrt{m^2\,-\,32m}}{8}\:\) [1]

We want the line to be tangent to the parabola,
\(\displaystyle \;\;\)so there will be exactly one intersection.
This happens when the discriminant is zero: \(\displaystyle \,m^2\,-\,32m\:=\:0\)

We have: \(\displaystyle \:m(m\,-\,32)\:=\0\;\;\Rightarrow\;\;m\:=\:0,\,32\)


Substitute \(\displaystyle m\,=\,0\) into [1]: \(\displaystyle \,x\:=\:0\)
\(\displaystyle \;\;\)Then: \(\displaystyle \,y\:=\:4\cdot0^2\:=\:0\)
The line is tangent at \(\displaystyle (0,0).\)


Substitute \(\displaystyle m\,=\,32\) into [1]: \(\displaystyle \,x\:=\:4\)
\(\displaystyle \;\;\)Then: \(\displaystyle \,y\:=\:4\cdot4^2\:=\:64.\)
The line is tangent at \(\displaystyle (4,64).\)

 

[galactus]

Here's a slightly different approach than Soroban's:

You started out correctly:

\(\displaystyle y-0=8x(x-2)\), but \(\displaystyle y=4x^{2}\)

\(\displaystyle 4x^{2}=8x(x-2)\)

You have a quadratic:

\(\displaystyle 4x^{2}-16x=0\)

\(\displaystyle 4x(x-4)=0\)

As you can see, x=0 and 4 are the solutions.

Using \(\displaystyle y=mx+b\ and\ y'=8x\)

If x=4, then 8(4)=32

\(\displaystyle 0=32(2)+b\)

b=-64

\(\displaystyle \H\\y=32x-64\)