I think y=2 there both on the 2 y intercept straight from each Other
Use distance formula to get a distance function in terms of x representing the distance from a point on curve (x,y) to point (3,0)
![d = \sqrt{(x-3)^2 + y^2](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B%28x-3%29%5E2%20%2B%20y%5E2)
y = 8/x
![d = \sqrt{(x-3)^2 + (\frac{8}{x})^2](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B%28x-3%29%5E2%20%2B%20%28%5Cfrac%7B8%7D%7Bx%7D%29%5E2)
Minimize the function by setting derivative equal to 0.
![\frac{d}{dx} = \frac{2(x-3) - 2(\frac{8}{x}) (\frac{8}{x^2})}{2 \sqrt{(x-3)^2 + (\frac{8}{x})^2}} = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%3D%20%5Cfrac%7B2%28x-3%29%20-%202%28%5Cfrac%7B8%7D%7Bx%7D%29%20%28%5Cfrac%7B8%7D%7Bx%5E2%7D%29%7D%7B2%20%20%5Csqrt%7B%28x-3%29%5E2%20%2B%20%28%5Cfrac%7B8%7D%7Bx%7D%29%5E2%7D%7D%20%3D%200)
Simplify and solve for x
![x-3 = \frac{64}{x^3}](https://tex.z-dn.net/?f=x-3%20%3D%20%5Cfrac%7B64%7D%7Bx%5E3%7D)
This becomes a 4th degree polynomial, you can use graphing calculator to find zeros.
Also intuitively you can recognize that 64 = 4^3, so if x=4 you get
1 = 1
Therefore x=4 is a solution, substituting give y =2
Answer:
did you ever find out?
Step-by-step explanation:
A. Because an expression with one or more terms is called a polynomial