Let p (x) = 4x^4-13x^2-2x
The zero of x-2 is 2
putting x = 2 in p (x), we get,
p (2) = 4×2^4-13×2^2-2×2
= 64 - 52 - 4
= 64 - 56
= 8
Therefore, remainder = 8

3 0

3 years ago

Two students in your class, Tucker and Karly, are disputing a function. Tucker says that for the function, between x = –3 and x

Effectus [21]

The difference between Tucker and Karly's take is that Tucker's solution is analytical while Karly's is graphical. But both are correct either way.

For Tucker's solution, let's say at x=-3 the value for y is 4, and at x=3, the value of y is still 4, then the average rate of change or slope is 0. Note that the slope of the curve is Δy/Δx. Since there is no change for Δy, the slope is zero.

For Karly's solution, even if the curve travels high or low but would have the same elevation of x=-3 and x=3, the average rate of change is still zero. It is actually just same with Tucker's but Karly just verbalizes her solution that was observed visually.

7 0

3 years ago