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anonymous
 one year ago
fan and medal
this is the equation f(x)=x^28x84
the vertex (4,100) axis of symmetery is 4
but how do I find three extra points to plot on the graph
anonymous
 one year ago
fan and medal this is the equation f(x)=x^28x84 the vertex (4,100) axis of symmetery is 4 but how do I find three extra points to plot on the graph

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, is a quadratic, so it'd look like a bowl so just pick one point to the left of the vertex another point to the right of the vertex and then you have the vertex, and draw the bowl so you really just need two, but if you need a third one, sure :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would do that but I need to show how I found tit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@pooja195 can you help please

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how you found it? well, you have the equation :), just pick any random "x" value and get a "y" and that's a point :)

pooja195
 one year ago
Best ResponseYou've already chosen the best response.0You're already getting help

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0if you factor the quadratic then you will have the 2 xvalues for when y = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would the other three points be (2,112) and (1,91)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@vera_ewing do you get this

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443588133366:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443588198587:dwdw:1443588313192:dw
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