Identify an equation in point-slope form for the line perpendicular to y = 3x + 5 that passes through (4, –1). a. y-4=-1/3(x+1) b. y+1=3(x-4) c. y-1=-3(x+4) d. y+1=-1/3(x-4)

the slope of the given formula y=3x+5 is 3

 a perpendicular line has a slope that is a negative reciprocal

 so that would be -1/3, so the answer needs to have -1/3 in it, that leaves you A and D

 the formula for point slope is y-y1=m(x-x1)

y1 = -1 and x1 = 4

 so you get: y+1=-1/3(x-4)

d 600 because 15*10*2=300

300+300=600

step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.