The problem states;
The graph of y=f(x) is given. Match each equation with its graph.
a) y=f(x-4)
b)y=2f(x+6)
c)y=f(x)+3
d)y=-f(2x)
What I find confusing is that these are parabolas. Why is this?
What are we matching these equations to? You don't have enough information here.
Given $\displaystyle y= f(x)$ is transformed to $\displaystyle y=a\times f(x-h)+k$
$\displaystyle a$ is a dilation factor away from the y axis, $\displaystyle h$ is a horizontal shift and $\displaystyle k$ a vertical shift.
sorry. Heres a link to a more descriptive analysis of the type of problem. Same instructions but different problem. It is problem number 4. The problem is in page 2.
http://www.farmingdale.edu/CampusPag..._Functions.pdf
http://www.farmingdale.edu/CampusPag..._Functions.pdf
The original graph of y=f(x) could be anything - in this case it is a parabola.
The question relates to transforming the original graph (whatever shape it is). There are basically 4 types of transformations:
y= a*f(x)
y=f(b*x)
y=f(x -c)
y=f(x) +d
The a, b, c, and d have different effects on the original graph of y=f(x).
Do you know what these effects are?
answer :
4 a) 2
b) ???
c) 1
d) 4
for 4 b) the equation is y = -f(x+4)
rearrange the equation become y = -f(x-(-4)) which is in standard form..
thus the graph of f(x) should be reflected across x-axis and shifted horizotally 4 unit to the left...
the graph for number 3 is -f(x+4)-1...
You need to understand what is going on here.
I think you probably typed in y=x-4 which IS a straight line, but that's irrelevant here.
I'll take you through the steps.
On your GC draw the parabola y=x^2 for starters. (You could do this with any graph but lets start with a basic parabola)
of course it will yield a straight line because you did nit define the function of f(x) is... since what you type is y= f(x-4) for sure it will sketch a straight line... try to the get the equation for the parabola shown in the figure... the substitute in the equation... the only you can sketch the parabola