## anonymous 4 years ago Create 2 parabolas with a smooth transition at P, out of the figure below. This is what I did: The first one I found using this technique. P1 = k(x+10)(x-10)+10 15 = k(0+10)(0-10) Removing 10 from y axis so I can treat the (10,10) as a zero (10,0). This gives P1 = -0.15x^2+25 P1 ' (10) = -3 --- P2 was trickier. I solved it like this: P2 = ax^2 + bx + c 100a + 10b + c = 10 P2 at 10 should be 10. 20a + b = -3 The derivative at 10 should be -3 b^2 = 4ac It has a single zero. I got the right solution but the method seems wrong to me. How would you solve it?

1. anonymous

2. anonymous

The answer is in the form: $\frac{9}{4} (x - \frac{50}{3})^{2}$

3. anonymous

9 / 40 *