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# Given are the equations f(x)= -3(x+1)^2 + 12 and g(x)=14-(√3-2x).

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Given are the equations f(x)= -3(x+1)^2 + 12 and g(x)=14-(√3-2x).

a. Give a window setting where you can read the points of intersections in the graph of g can read with the axis (I use a TI-8 Plus CE-T graphing calculator and normally use the zstandard option to fit my graph, but I'm not really sure if it shows everything of the graph)

b. Calculate the coordinates of the points of intersections of both graphs. Give the answers rounded to two decimals places

Feb 20, 2020

#1
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Well....I don't know if this helps  but.....here's a graph with  the intersection points shown :

https://www.desmos.com/calculator/o3gydmyzdf   Feb 20, 2020
#2
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Ah I'm sorry I placed the brackets weirdly for g(x)... it's meant to be √(3-2x). I did get the window setting alright now but could you maybe help me with b. ? I'm not really sure how to calculate the coordinates

Guest Feb 20, 2020
#3
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We have

y  = -3 ( x + 1)^2  + 12

y = 14  - √(3 -2x]

Setting these  equal  we have that

-3 ( x + 1)^2  + 12  =  14 - √[3-2x]

-3x^2 -6x -3  + 12  = 14  - √[3-2x]

-3x^2 -6x + 9  = 14 - √[3-2x]

-3x^2  - 6x - 5  =  -√[3-2x]        multiply through by    -1

3x^2 + 6x + 5  = √[3-2x]

This  would be very dfficult to solve algebraically

Here's  a graph  of  both sides of the equation with the intersection points shown :

https://www.desmos.com/calculator/fsjl5mhaqo

Remember to round these to two decimal places....   Feb 20, 2020
edited by CPhill  Feb 20, 2020
#4
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Thank you! So solving it algebraically would give me the points of intersection? The assignments asks for me to show my calculation but if it is that difficult i might just skip it then.

Guest Feb 20, 2020
#5
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The calculation of these solutions algebraically is above my pay-grade.....LOL!!!!

WolframAlpha   shows  the  exact  solutions here :

https://www.wolframalpha.com/input/?i=-3+%28+x+%2B+1%29%5E2++%2B+12++%3D++14+-+%E2%88%9A%5B3-2x%5D

As you can see.....these  are  really "messy "

A graph seems the way to go.....!!!!   CPhill  Feb 20, 2020
#6
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Haha I see! Either way thanks for the help. I really appreciate it!! Have a good day

Guest Feb 20, 2020