Problem 1: Find an equation for the line with y-intercept 3 that is perpendicular to the line y=2/3x-4. & Problem 2: If P(4,-5) is a point on the graph of the function y=f(x), find the corresponding point on the graph of y=2f(x-6). & Problem 3: The polynomial f(x) divided x-3 results in a quotient of x^2+3x-5 with a remainder of 2. Find f(3).
2006-11-20 13:51:00 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
the first one is
y= -3/2+3
2006-11-20 13:54:33 · answer #1 · answered by pirate_ninjarawr 2 · 0⤊ 0⤋
y= 3/2 +3
2006-11-20 21:59:08 · answer #2 · answered by Kate F 1 · 0⤊ 0⤋
Problem 1:
The slope of y=2/3x-4 is 2/3.
Therefore the slope of a line perpendiular to it must be -3/2
Therefore the equation desired is y=(-3/2)x +3
Problem 2:
P(4,5) is a point on y=f(x)
Same point on y=2f(x-6) would double y and move x 6 places to the left giving the new point (-2,10)
Problem3: By the Remainder Theorem f(3) =2
2006-11-20 22:26:24 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
1. y= -3/2x + 3
2006-11-20 21:54:23 · answer #4 · answered by Metaspy 3 · 0⤊ 0⤋
1. Whatever you add or subtract from x = the y intercept,hence the answer is y=2/3x+4
Sorry....I don't know the rest (i'm only in 8th grade)
2006-11-20 21:53:27 · answer #5 · answered by CIA Biatch 3 · 0⤊ 2⤋
2! duh everyone knows that! no srry i would help you but i have an F in math :D
2006-11-20 21:54:37 · answer #6 · answered by i_love_black13 1 · 0⤊ 0⤋