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anonymous
 one year ago
Find a quadratic model for the set of values: (2, 20), (0, 4), (4,20) Show your work
anonymous
 one year ago
Find a quadratic model for the set of values: (2, 20), (0, 4), (4,20) Show your work

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perl
 one year ago
Best ResponseYou've already chosen the best response.2You need a system of equations

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I don't feel well today. I woke up with a migraine (again)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Take some advil and drink water

perl
 one year ago
Best ResponseYou've already chosen the best response.2$$ \Large{ y = ax^2 + bx + c \\~\\ 20 = a(2)^2 + b(2) + c \\4 = a(0)^2 + b(0) + c \\ 20 = a(4)^2 + b(4) + c }$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i honestly have no clue how to do this equation

perl
 one year ago
Best ResponseYou've already chosen the best response.2Do you agree with what I have so far. I plugged in the values into a generic quadratic function.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think so, i wouldn't know if you were wrong

perl
 one year ago
Best ResponseYou've already chosen the best response.2Finding a quadratic model means fitting the points to a quadratic curve \( \large y = ax^2 + bx + c \) . So our job is to find the parameters \( \large a,b,c\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@perl how do i find the a,b,c

perl
 one year ago
Best ResponseYou've already chosen the best response.2We have these three equations so far. \[ \\~\\\Large { 20 = a(2)^2 + b(2) + c \\4 = a(0)^2 + b(0) + c \\ 20 = a(4)^2 + b(4) + c }\]

perl
 one year ago
Best ResponseYou've already chosen the best response.2the second equation gives us 4 = 0 + 0 + c 4 = c

perl
 one year ago
Best ResponseYou've already chosen the best response.2Yes. Now back substitute.

perl
 one year ago
Best ResponseYou've already chosen the best response.2We have these three equations so far. \[ \\~\\\Large { 20 = a(2)^2 + b(2) + (4) \\4 = a(0)^2 + b(0) + (4) \\ 20 = a(4)^2 + b(4) + (4) }\]

perl
 one year ago
Best ResponseYou've already chosen the best response.2we don't need the second equation any longer

perl
 one year ago
Best ResponseYou've already chosen the best response.2\[ \\~\\\Large { 20 = a(2)^2 + b(2) + (4) \\ 20 = a(4)^2 + b(4) + (4) }\]

perl
 one year ago
Best ResponseYou've already chosen the best response.2\[ \\~\\\Large { 20 = a(2)^2 + b(2) 4 \\ 20 = a(4)^2 + b(4) 4 \\~\\} \rm simplify \\~\\\Large { 20 = 16a 2b 4 \\ 20 = 16a + 4b 4 \\~\\ \rm simplify ~ more\\~\\ 16 = 16a 2b \\ 16 = 16a + 4b }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that the final answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we haven't solved A or C yet?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so how do we solve for A and B than? Im sorry i really dont understand

perl
 one year ago
Best ResponseYou've already chosen the best response.2we can solve that linear equation

perl
 one year ago
Best ResponseYou've already chosen the best response.2\[ \\~\\\Large { 20 = a(2)^2 + b(2) 4 \\ 20 = a(4)^2 + b(4) 4 \\~\\} \rm simplify \\~\\\Large { 20 = 16a 2b 4 \\ 20 = 16a + 4b 4 \\~\\ \rm simplify ~ more\\~\\ 16 = 16a 2b \\ 16 = 16a + 4b \\~\\ \rm add ~ the ~ two ~ equations\\~\\ \\32 = 2b \\~\\ \\b = \frac{32}{2} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So now we need to find A

perl
 one year ago
Best ResponseYou've already chosen the best response.2y = 2x^2 + 4x 4 should be the answer. I have an error somewhere above.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so b should really be b=4?

perl
 one year ago
Best ResponseYou've already chosen the best response.2I have to go back and find the error now ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome thank you so much!

perl
 one year ago
Best ResponseYou've already chosen the best response.2but this method is sound if you want to try it again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i prob will, i have a whole other algebra class ahead of me

perl
 one year ago
Best ResponseYou've already chosen the best response.2\[ \\~\\\Large { 20 = a(2)^2 + b(2) 4 \\ 20 = a(4)^2 + b(4) 4 \\~\\} \rm simplify \\~\\\Large { 20 = \color{red}4a 2b 4 \\ 20 = 16a + 4b 4 \\~\\ \rm simplify ~ more\\~\\ 16 = \color{red}4a 2b \\ 16 = 16a + 4b \\~\\ \rm solve ~this~ system~ of~ equations\\~\\ \\ a = 2 ~ , b = 4 \\~\\ \rm backsubstitute \\y = 2x^2 + 4x  4 }\]

perl
 one year ago
Best ResponseYou've already chosen the best response.2Another way to solve this problem is to note that your parabola has a y intercept of 4, because x=0 at the y intercept. That reduces the problem of solving a linear system of 2 equations as we have above.
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