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MakcuM [25]
3 years ago
14

Graph y = x2 + 2. Identify the vertex of the graph. Tell whether it is a minimum or maximum

Mathematics
1 answer:
Leona [35]3 years ago
4 0
Vertex is -b/2a. The variable b=2 and a=2. That finds the x coordinate of the vertex. Then to find the y coordinate you just plug the number you got for x (-1/2) into the equation. Like so:
y = 2x+2
y = 2(-1/2)+2
y = -1+2
y = 1
So your vertex is ( -1/2, 1 )
To graph your vertex you take your x coordinate (-1/2) and find it on the x axis of the graph (horizontal line). Then do the same for y (1) (on the vertical axis.
Now that you have your vertex graphed you can find out if it is minimum or maximum by looking at the direction in which the slope is going. If it goes in the positive direction (up) then it is a minimum since that is the lowest point that the line touches. If it is maximum it is opposite. The line faces down and the highest point the line touches is that point.
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Answer:

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you will need to find the distance between each segment-

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d=\sqrt{(x_{2}-x_{1})  ^{2}+(y_{2}-y_{1})  ^{2}  }

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d=\sqrt{(-1--2)^{2} +(2--1)^{2} } \\d=\sqrt{1^{2}+3^{2}  } \\d=\sqrt{10} =3.16

CF:

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d=\sqrt{(1--1)^{2} +(1-2)^{2} } \\d=\sqrt{2^{2}+-1^{2}  } \\d=\sqrt{4+1} \\d=\sqrt{x} 5=2.24

FE

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d=\sqrt{(2-1)^{2} +(-2-1)^{2} } \\d=\sqrt{1^{2} +(-3)^{2} } \\d=\sqrt{1+9} \\d=\sqrt{10}= 3.16

ED

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d=\sqrt{(-2-2)^{2} +(-1--2)^{2} } \\d=\sqrt{(-4)^{2}+1^{2}  } \\d=\sqrt{16+1} \\d=\sqrt{17}= 4.12\\

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hope that helps!



4 0
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