Remember, h(x) = y and x ...well, x is x...lol
B. h(x) = |x|....subbing in (-1,1)
y = |x|
1 = |-1|...absolute values are positive, even when they say negative
1 = 1 (correct)
The polynomial whose zeroes are 1, 2 and 3 is given by,
(x-1)(x-2)(x-3) = 0
(x-1)[x²-3x-2x+6]=0
(x-1)[x²-5x+6]=0
x[x²-5x+6] -1[x²-5x+6]=0
x³-5x²+6x-x²+5x-6=0
x³-6x²+11x-6=0
Therefore, the required polynomial is,
x³ - 6x² + 11x -6 = 0
Answer:
use a website called symbolab it give you the ansewrs and an explanation
Step-by-step explanation:
Are you talking about a box and whisker plot??? If you are, here's some notes I took when I learned that :) sorry about my messy hand writing. Hope this helps.