+2498 \(if\:\:m\ne 0,\:m\ne 1, and\:\;f(x)=mx+b,\:\text{then which of}\\ \text{the following statements concering the graphs whose}\\ \text{are} \;y=f(x)+3\;\;and\;y=f(x+3)\;\text{must be true?}\)
(A) the graph don't intersect
(B) the graphThe graph intersect in one point
(C) the graph intersect in two points
(D) the graphin more than two points
I get
\(\begin{array}{rcll} m(x-3)+b &=&mx+b+3 \\ mx-3m &=& mx+3\\ m&=&-1 \end{array}\)
I think D because their slopes are the same but (x-3) is tricking
+23252 Let's take a particular example; let's say that f(x) = 7x + 29.
y = f(x) + 3 = (7x + 29) + 3 = 7x + 32
y = f(x + 3) = 7(x + 3) + 29 = 7x + 21 + 29 = 7x + 50
This shows that, for this case, the two lines are parallel because they have the same slope but different y-intercepts.
This says that the answer is (A).
f(x) + 3 takes the original function, f(x), and moves it up 3 spaces but leaves the slope alone.
f(x + 3) makes the y-value of f(x) the y-value of f(x + 3); it moves it 3 spaces down the line but leaves the slope alone.
My suggestion to doing problems like this: take a particular example but don't use values for m and b that are special numbers, such as 0, 1 or -1. Then, analyze what happens to the particular beginning example.
+23252 Let's take a particular example; let's say that f(x) = 7x + 29.
y = f(x) + 3 = (7x + 29) + 3 = 7x + 32
y = f(x + 3) = 7(x + 3) + 29 = 7x + 21 + 29 = 7x + 50
This shows that, for this case, the two lines are parallel because they have the same slope but different y-intercepts.
This says that the answer is (A).
f(x) + 3 takes the original function, f(x), and moves it up 3 spaces but leaves the slope alone.
f(x + 3) makes the y-value of f(x) the y-value of f(x + 3); it moves it 3 spaces down the line but leaves the slope alone.
My suggestion to doing problems like this: take a particular example but don't use values for m and b that are special numbers, such as 0, 1 or -1. Then, analyze what happens to the particular beginning example.
