Can somebody answer this ?? :)
2 answers:
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if we zero out f(x), namely make y = 0, we can get the roots or x-intercepts for this quadratic equation
now, the equation is in x-terms, meaning is a vertically opening parabola, so the axis of symmetry will be x = something, a vertical line.
well, we have two x-intercepts, one at -4 and another at 2, and the vertex is right half-way between those guys
-4------------(-1)------------2
so the vertex is at x=-1, namely the axis of symmetry is x = -1.
The unit disk can be parameterized by the function
where
and 
. The squared distance between any point in this region 
and the point (1, 1) is
The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.
![=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7D%5Br%5E2-2r%28%5Ccos%5Ctheta-%5Csin%5Ctheta%29%2B2%5Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D)
where
The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.