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