Question 675159
Let f(x) be the parabola -x^2+16x-16. Find the point (x,y) on f such that tangent line to f at (x,y) passes through the origin.
<pre>
The way to approach this depends on whether you are taking algebra or
taking calculus.  I will assume you are taking algebra.  If you are
taking calculus, tell me in the thank you note and I'll help you
using calculus.

If you are taking algebra.  You need this fact:

The graph of a line g(x) is tangent to the graph of parabola f(x) 
if and only if the equation g(x)=f(x) has exactly one solution.

Let the line that is to be tangent to the parabola f(x) have the 
equation g(x) = mx + b.  Since g(x) is to pass through the origin
(x,y) = (0,0) must satisfy that equation:

              g(x) = mx + b
                 0 = m·0 + b
                 0 = b

So            g(x) = mx + 0
              g(x) = mx

Now we form the equation:

              g(x) = f(x)
                mx = -x² + 16x - 16
x² + mx - 16x + 16 = 0
 
 x² + (m-16)x + 16 = 0

That equation will have exactly one solution if its
discriminant B²-4AC = 0, where A=1, B = m-16, C=16

discriminant = B²-4AC = (m-16)²-4(1)(16) = (m-16)²-64

We set that discriminant = 0

     (m-16)² - 64 = 0
          (m-16)² = 64
Use the principle of square roots:

             m-16 = ±8

                m = 16±8

There are two solutions for m, 16+8=24 and 16-8=8

Using m = 24

 x² + (m-16)x + 16 = 0 
 x² + (24-16)x + 16 = 0
       x² + 8x + 16 = 0
           (x + 4)² = 0
                  x = -4
The line's equation is g(x) = mx
                       g(x) = 24x
                      g(-4) = 24(-4)
                      g(-4) = -96 
So the point of tangency is (-4,-96)

The graph is:

{{{drawing(400,400,-10,20,-200,100, graph(400,400,-10,20,-200,100,-x^2+16x-16),
 graph(400,400,-10,20,-200,100,24x), locate(-9,-90,"(-4,-96)"),
circle(-4,-96,.3) )}}}

There is another solution:

Using m = 8

 x² + (m-16)x + 16 = 0 
 x² + (8-16)x + 16 = 0
       x² - 8x + 16 = 0
           (x - 4)² = 0
                  x = 4
The line's equation is g(x) = mx
                       g(x) = 8x
                      g(4) = 8(4)
                      g(4) = 32 
So the point of tangency is (4,32)

The graph is:

{{{drawing(400,400,-10,20,-200,100, graph(400,400,-10,20,-200,100,-x^2+16x-16),
 graph(400,400,-10,20,-200,100,24x), locate(-9,-90,"(-4,-96)"),
circle(-4,-96,.3),   graph(400,400,-10,20,-200,100,8x),locate(4,32,"(4,32)"),
circle(4,32,.3)

 )}}}

Edwin</pre>