Since we can find that the graphs Y value is by a factor of 2 looking at the graph we can find
f(3)=8
(x-h)²+(y-k)²=r² is the equation of a circle with radius of r
so if x-h=4 and y-k=3 then
4²+3²=r²
16+9=r²
25=r²
sqrt both sides
5=r
radius is 5 units
Check the picture below, so the parabola looks more or less like that.
now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.
![\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Ctextit%7Bhorizontal%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bfocus~point%7D%7B%28h%2Bp%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7Bx%3Dh-p%7D%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22p%22~is~negative%7D%7Bop%20ens~%5Csupset%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Csubset%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
![\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20h%3D-7%5C%5C%20k%3D-2%5C%5C%20p%3D-4%20%5Cend%7Bcases%7D%5Cimplies%204%28-4%29%5Bx-%28-7%29%5D~~%20%3D%20~~%5By-%28-2%29%5D%5E2%20%5C%5C%5C%5C%5C%5C%20-16%28x%2B7%29%3D%28y%2B2%29%5E2%5Cimplies%20x%2B7%3D-%5Ccfrac%7B%28y%2B2%29%5E2%7D%7B16%7D%5Cimplies%20x%3D-%5Ccfrac%7B1%7D%7B16%7D%28y%2B2%29%5E2-7)
The equation that can be used to find the coldest temperature recorded in Siberia is y + 10 = -80
Let the coldest temperature in Alaska be x
The coldest temperature recorded in Alaska was –80°F
x = –80°F
Let the coldest temperature in Siberia be y
The coldest temperature recorded in Alaska was 10°F warmer than the coldest temperature recorded in Siberia
This can be interpreted as:
Coldest temperature in Alaska = Coldest temperature in Siberia + 10
x = y + 10................(*)
Since x = -80
Substitute x = -80 into equation (*)
-80 = y + 10
This can be re-written as:
y + 10 = -80
The equation that can be used to find the coldest temperature recorded in Siberia is y + 10 = -80
Learn more here: brainly.com/question/2159099
