Help, quickly!

Question

0

702

1

+5265

Please answer in something at least somewhat understandable. Thank you! Also, please try to keep it related to the question.

Melody · Answer 1 · 2016-10-25T08:22:58

Hi rarinstraw,

Gino is really good at these he may have a better way to do it.

I set up these four equations

$cost = ax^2+bx+c\\~\\ 550=a*225+b*15+c\\ 700=a*289+b*17+c\\ 1500=a*529+b*23+c\\ 2500=a*900+b*30+c$

Then I solved them simultaneously.

I got

$a\approx 3.23\\ b\approx -28.40\\ c\approx 249\\ so\\ cost\approx 3.23x^2-28.40x+249$

3.23*42^2-28.40*42+249 = 4753.92

The points do not fit on the graph exactly because the these points will not fit any quadratic model exactly.

So the estimated cost of a 42cm tele is $4754

BUT since 42cm is an extrapolated point (larger than any of the other sizes) you would need to be very careful about relying on this price.