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A portion of the graph of $f(x)=ax^3+bx^2+cx+d$ is shown below. What is the value of $8a-4b+2c-d$?

Guest Jun 17, 2017
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 #1
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This is the graph of a cubic equation that has zeros at 0, 1, and 2.

Its function is:  y  =  a(x - 0)(x - 1)(x - 2)

[There is no constant term since the graph passes through the origin.]

               

Since this function passes through the point (4, 3), we can substitute 4 for x and 3 for y:

     y  =  a(x - 0)(x - 1)(x - 2)     --->     3  =  a(4 - 0)(4 - 1)(4 - 2)     --->     3  =  a(4)(3)(2)     --->     3  =  24a

              --->  a = 1/8

 

Therefore, the function is:  y  =  (1/8)(x)(x - 1)(x - 2)     --->     y  =  (1/8)(x3 - 3x2 + 2x)

               --->      y  =  (1/8)x3 - (3/8)x2 + (1/4)x

 

--->     a = (1/8)     b = -3/8     c = 1/4     d = 0

 

--->     8a - 4b + 2c - d  =  8(1/8) -4(-3/8) + 2(1/4) - 0     =     1 + 3/2 + 1/2 - 0     =     3

geno3141  Jun 17, 2017

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