# Thread: Graph of the Function

1. ## Graph of the Function

Graph the function f below by starting with the graph of y = x^2 and using transformations (shifting, compressing, stretching and/or reflection.

HINT: If needed, write f in the form f(x) = a(x - h)^2 + k.

f(x) = (2/3)x^2 + (4/3)(x) - 1

I need someone to show me how to graph such a function f.

2. Originally Posted by symmetry
Graph the function f below by starting with the graph of y = x^2 and using transformations (shifting, compressing, stretching and/or reflection.

HINT: If needed, write f in the form f(x) = a(x - h)^2 + k.

f(x) = (2/3)x^2 + (4/3)(x) - 1

I need someone to show me how to graph such a function f.
Hello,

first transcribe the equation of your function into the given general form:

$f(x)=\frac{2}{3}x^2+\frac{4}{3}x-1$
$f(x)=\frac{2}{3}(x^2+2x)-1$
$f(x)=\frac{2}{3}((x^2+2x+1)-1)-1$
$f(x)=\frac{2}{3}(x+1)^2-\frac{2}{3}-1$
$f(x)=\frac{2}{3}(x+1)^2-\frac{5}{3}$

Now you can describe the steps of transformation of the normal parabola:

1. shift 1 unit to the left because of (x+1)
2. compress the graph by the factor $\frac{2}{3}$ because the factor is smaller than 1
3. move down by $\frac{5}{3}$

I've attached a diagram which contains these three steps. The order of transformation is: black - blue - green - red

EB

3. ## ok

What can I say? I am overwhelmed by your kindness and willingness to help me with tough math questions.

4. Originally Posted by earboth
Now you can describe the steps of transformation of the normal parabola:

1. shift 1 unit to the left because of (x+1)
2. compress the graph by the factor $\frac{2}{3}$ because the factor is smaller than 1
I know what you are saying here, but theoretically you need to compress the graph first, then shift 1 unit to the left. Since you are compressing the entire x-axis the location of the vertex would also shift, unless the vertex is at x = 0. (In other words the way you have done it puts the vertex at x = -2/3, not x = -1.)

-Dan

5. Originally Posted by topsquark
... Since you are compressing the entire x-axis the location of the vertex would also shift, unless ...
-Dan
Hello, Dan,

The factor $\frac{2}{3}$ scales down only all y-values of coordinates of points. Geometrically it is a dilation(?) of the complete coordinate-plane perpendicular to the x-axis. All points forming the x-axis are fixed points(?) and stay therefore unmoved.