can someone help me with this question is super confusing.
cube with side length x has been modified by adjusting its side length x has been modified by adjusting its length, width and height. A student determined a function of the form v(x)=(x+a)(x+b)(x+c) can be used to represent the volume of the new box for different values of x, and then she graphed the function.
a.based on the graph ,predict how the side lengths of the original cube changed.(hint:what do the zeros of the function represent?
b.determine a polynomial function of the form v(x)=(x+a)(x+b)(x+c) that's matches the function in the graph .
c.which parts of the graph are realistic?(hint:all sides must be positive.using two negative side lengths to give positive volume is not realistic.)
4 years ago
Answered By Prabhat K
(a,b) From the zeros of the graph we can determine the change of the side lengths. The zeros of this graph are: -1,+2,+4. In terms of the function, we have V(x)=(x+1)(x-2)(x-4). Therefore the side lengths have changed by +1,-2,-4.
(C) the conditions must be that the factors (x+a) must be greater than 0. To ensure the modified side length is realistic.
4 years ago
Answered By Prabhat K
Adding to my previous response, for part c, you cannot have two negative factors.
4 years ago
Answered By Albert S
The zeros of the function allow us to determine the factors of the expression which lead us to a, b, c.
I read the graph to have zero crossings of x=(-1) x = (2) and x = (4)
These values of x must have come from solving the expressions (x+1)=0, (x-2)=0 and (x-4)=0
V(x) = (x+1)(x-2)(x-4)
The meaning of these factors is two sides get smaller than the original side of the cube, one side gets larger than the original cube.
It is physically impossible to have a side length smaller than zero (a negative side length has no meaning and a side length of zero will not enclose any space - these are just the mathematica extremes).
The portion of the gragh with negative x is not an expression of reality because the original cube also cannot have a negative length.
Looking at the three factors;
the factor (x+1) will never become negative for positive values of x,
the first factor to go negative will be (x-4) because you are removing a larger piece from that side. As a reference if x had started as 5 (the cube would have had a volume of 5x5x5=125) - the new dimensions would be (6) (3) (1) giving a volume of 6x3x1=18 (which looks like where the graph is going). If the original cube had a side length of 4 - the new dimensions would be (5)(2)(0) these side lengths will not form a solid object.
Therefore the only usable portion of the grapg is for the region x>4