The required function is, f(x) = 2x³+3x²-x-4
The general equation of cubic regression is,
f(x) = ax³+bx²+cx+d , where a,b,c,d are real numbers which are co-efficients of the variable terms.
The point (0, -4) indicates that y-intercept is -4,
So the cubic regression equation becomes,
f(x) = ax³+bx²+cx-4
Another three given points are (-3,-28), (4,168) & (-5,-174)
Now, for the 1st point,
f(-3) = a(-3)³+b(-3)²+c(-3)-4
⇒ -28 = -27a+9b-3c-4
⇒ 27a-9b+3c = 24 ........(1)
For the 2nd point,
f(4) = a(4)³+b(4)²+c(4)-4
⇒ 168 = 64a+16b+4c-4
⇒ 64a+16b+4c = 172 ........(2)
For the 3rd point,
f(-5) = a(-5)³+b(-5)²+c(-5)-4
⇒ -174 = -125a+25b-5c-4
⇒ 125a-25b+5c = 170 .........(3)
Now multiply 27 with (2) & multiply 64 with (1), then subtract (1) from (2),
27(64a+16b+4c)-64(27a-9b+3c) = 4644-1536
⇒ 1728a+432b+108c-1728a+576b-192c = 3108
⇒ 1008b-84c = 3108
⇒ 84b-7c=259 ......(4)
Now multiply 125 with (2) & multiply 64 wuth (3), then subtract (3) from (2),
125(64a+16b+4c)-64(125a-25b+5c) = 21500-10880
⇒ 8000a+2000b+500c-8000a+1600b-320c = 10620
⇒ 3600b+180c = 10620
⇒ 20b+c=59 ........(5)
So, multiply 20 with (4) & multiply 84 with (5), then subtract (5) from (4),
20(84b-7c)-84(20b+c)=5180-4956
⇒ 1680b-140c-1680b-84c=224
⇒ 224c = -224
⇒ c = -1
From (5),
20b-1=59
⇒ 20b = 60
⇒ b = 3
Now, from (1),
27a-9(3)+3(-1)=24
⇒ 27a-27-3=24
⇒ 27a -30 = 24
⇒ 27a = 54
⇒ a = 2
So, a=2, b=3 & c=-1
The equation becomes, f(x) = 2x³+3x²-x-4
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