Choice A) The vertical line test confirms that the graph is a function
True. It is impossible to pass a single vertical line through more than one point on this graph.
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Choice B) The domain is x >= 2
False. The domain is the set of all real numbers
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Choice C) The range is y >= 2
True. The smallest y value possible is y = 2. This is where the lowest part of the graph occurs.
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Choice D) This piecewise function is made up of four intervals
False. There are two intervals (the curved one on the left; the straight piece on the right)
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Choice E) Each interval of the piecewise function contains a graph of a linear function.
False. While the piece on the right is linear, the left portion is nonlinear (curved piece).
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In summary we have:
choice A) true
choice B) false
choice C) true
choice D) false
choice E) false
So the answers are choice A and choice C
Check the picture below. So the parabola looks more or less like so.
![\bf \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] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%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~%5Cdotfill)
![\bf \begin{cases} h=-5\\ k=2\\ p=4 \end{cases}\implies 4(4)[x-(-5)]=[y-2]^2\implies 16(x+5)=(y-2)^2 \\\\\\ x+5=\cfrac{1}{16}(y-2)^2\implies x = \cfrac{1}{16}(y-2)^2-5](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20h%3D-5%5C%5C%20k%3D2%5C%5C%20p%3D4%20%5Cend%7Bcases%7D%5Cimplies%204%284%29%5Bx-%28-5%29%5D%3D%5By-2%5D%5E2%5Cimplies%2016%28x%2B5%29%3D%28y-2%29%5E2%20%5C%5C%5C%5C%5C%5C%20x%2B5%3D%5Ccfrac%7B1%7D%7B16%7D%28y-2%29%5E2%5Cimplies%20x%20%3D%20%5Ccfrac%7B1%7D%7B16%7D%28y-2%29%5E2-5)
Step-by-step explanation:
working is above
hope it helps
The end of the bar should end in the exact middle of 8 and 10.
Hope this helps! :)
The answer is 21 because you do PEMDAS first. (8-8) is 0. And 0+21 is 21
