# Graphing

• May 23rd 2009, 06:08 PM
foreverbrokenpromises
Graphing
The original graphy, $y=f(x)$ is transformed into a new graph $y=4 f(2x-2)-3$ using progression. If the point (4, -11) is on the transformed graph, then a point guranteed to be on the original graph is :
a) 3, -1
b) 6, -16
c) 4, -2
d) 6, -2
e)1, -2

my teacher gave me a package of 50 questions and didnt explain to us how to do the questions. he just told us to do them
so i am very confused right now!
• May 23rd 2009, 06:31 PM
skeeter
Quote:

Originally Posted by foreverbrokenpromises
The original graphy, $y=f(x)$ is transformed into a new graph $y=4 f(2x-2)-3$ using progression. If the point (4, -11) is on the transformed graph, then a point guranteed to be on the original graph is :
a) 3, -1
b) 6, -16
c) 4, -2
d) 6, -2
e)1, -2

my teacher gave me a package of 50 questions and didnt explain to us how to do the questions. he just told us to do them
so i am very confused right now!

$y = 4 \cdot f(2x-2) - 3$

sub in the values from the given point for the transformation ... $x = 4$ , $y = -11$

$-11 = 4 \cdot f(2(4) - 2) - 3$

simplify ...

$-2 = f(6)$