anonymous
  • anonymous
Find a quadratic model for the set of values: (-2, -20), (0, -4), (4,-20) Show your work
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
perl
  • perl
You need a system of equations
UsukiDoll
  • UsukiDoll
I don't feel well today. I woke up with a migraine (again)

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anonymous
  • anonymous
Take some advil and drink water
perl
  • perl
$$ \Large{ y = ax^2 + bx + c \\~\\ -20 = a(-2)^2 + b(-2) + c \\-4 = a(0)^2 + b(0) + c \\ -20 = a(4)^2 + b(4) + c }$$
anonymous
  • anonymous
i honestly have no clue how to do this equation
perl
  • perl
Do you agree with what I have so far. I plugged in the values into a generic quadratic function.
anonymous
  • anonymous
I think so, i wouldn't know if you were wrong
perl
  • perl
Finding 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
  • anonymous
@perl how do i find the a,b,c
perl
  • perl
We 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
  • perl
the second equation gives us -4 = 0 + 0 + c -4 = c
anonymous
  • anonymous
So c=-4
perl
  • perl
Yes. Now back substitute.
perl
  • perl
We 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
  • perl
we don't need the second equation any longer
perl
  • perl
\[ \\~\\\Large { -20 = a(-2)^2 + b(-2) + (-4) \\ -20 = a(4)^2 + b(4) + (-4) }\]
anonymous
  • anonymous
im writing as we go
perl
  • perl
\[ \\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = 16a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = 16a -2b \\ -16 = 16a + 4b }\]
anonymous
  • anonymous
Is that the final answer?
perl
  • perl
We can solve for `b`
anonymous
  • anonymous
So we haven't solved A or C yet?
perl
  • perl
we solved for c = -4
anonymous
  • anonymous
Ok so how do we solve for A and B than? Im sorry i really dont understand
anonymous
  • anonymous
perl
  • perl
we can solve that linear equation
anonymous
  • anonymous
how?
perl
  • perl
\[ \\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = 16a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = 16a -2b \\ -16 = 16a + 4b \\~\\ \rm add ~ the ~ two ~ equations\\~\\ \\-32 = 2b \\~\\ \\b = \frac{-32}{2} }\]
anonymous
  • anonymous
So b=-16?
perl
  • perl
yes
anonymous
  • anonymous
So now we need to find A
perl
  • perl
y = -2x^2 + 4x -4 should be the answer. I have an error somewhere above.
anonymous
  • anonymous
so b should really be b=4?
anonymous
  • anonymous
A=-2 B=4 C=-4?
perl
  • perl
yes
perl
  • perl
I have to go back and find the error now ;)
anonymous
  • anonymous
Awesome thank you so much!
perl
  • perl
but this method is sound if you want to try it again
perl
  • perl
if you ever*
anonymous
  • anonymous
yeah i prob will, i have a whole other algebra class ahead of me
perl
  • perl
\[ \\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = \color{red}4a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = \color{red}4a -2b \\ -16 = 16a + 4b \\~\\ \rm solve ~this~ system~ of~ equations\\~\\ \\ a = -2 ~ , b = 4 \\~\\ \rm backsubstitute \\y = -2x^2 + 4x - 4 }\]
perl
  • perl
Another 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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