y=f(x+5/2)
I timesd x+5 by 2 to get rid of 2 at the bottom and timesd y by 2. Then to get rid of 2 from y i divided it, and i got this.
y=1/2f(x+5)
however my teacher got y=f(1/2(x+5))
what did i do wrong?
The problem seems to be ambiguous notation.
Without more parentheses, it is unclear whether your
y=f(x+5/2) is
y = f[(x+5)/2]
or
y = f[x + (5/2)]
They are not the same thing.
One of those agrees with your teacher's
f[(1/2)(x+5)]
The question in my book only has brackets around (x+5/2)
I still don't understand the whole process
Your approach to manipulating the equation is incorrect. Let's walk through it step by step to see where the mistake lies.
Given: y = f(x + 5/2)
Your incorrect manipulation:
1. You multiplied both sides of the equation by 2 to get: 2y = 2f(x + 5/2)
2. Then, you divided both sides of the equation by 2 to get: y = f(x + 5/2)
The issue with your approach is that when you multiplied both sides of the equation by 2, you actually distributed the 2 to both terms inside the parentheses (x and 5/2). This means that you should have multiplied x by 2 and 5/2 by 2 as well.
Correct manipulation of the equation using distribution:
1. Multiply both terms inside the parentheses by 2: y = f(2x + 10/2)
2. Simplify the expression in parentheses: y = f(2x + 5)
As you can see, your teacher's answer, y = f(1/2(x + 5)), is correct. The original function, y = f(x + 5/2), can be rewritten as y = f(1/2(x + 5)) when using correct algebraic manipulation.