MATH533 Ashford University Week 5 Vertical Line Test Discussion

Graph each relation, and state its domain and range

Sketch the graph of each function, and state the domain and range. See Examples 5–7. See the Strategy for graphing y = af(x − h) + k

<

WEEK 5 DISCUSSION PROBLEMS FOR ALGEBRAYOU MUST OPEN THE ATTACHMENT TO SEE THE PROBLEMS BEFORE
BIDDING THE LAST 4 WEEKS I HAVE NOT BEEN GETTING FULL POINTS ON
THIS MATH STUFF BECAUSE DIRECTIONS HAVEN’T BEEN FOLLOWED
Your initial post should be at least 250 words in length.
For each of the above problems, do the following:
•
•
•
•
•
•
•
For each of your equation, write down five (5) points on the graph of each equation (you
do not have to submit your graph. Draw the graph using paper/pencil or use online
graphing calculators like Desmos Graphing Calculator (Links to an external site.) to
answer the question).
For each of your equation, does the graph of each equation have any intercepts? If yes,
what are they? If not, why not?
State the domain for each of your equations. Write them in interval notation.
State the range for each of your equations. Write them in interval notation.
State whether each of the equations is a function or not giving your reasons for the
answer.
Select one of your graphs and assume it has been shifted three units upward and four
units to the left. Discuss how this transformation affects the equation by rewriting the
equation to incorporate those numbers.
Incorporate the following four math vocabulary words into your discussion. Use bold
font to emphasize the words in your writing. Do not write definitions for the words; use
them appropriately in sentences describing the thought behind your math work.
o Function
o Relation
o Vertical Line test
o Transformation
OKfinal week aka week 5 #45.
My first equation is y=-3+6.5. First things first we need to plot 5 points,
XY
-58
0 6.5
16.2
55
21.60
The next part is asking if the line equation has any intercepts. As my above plot shows this equation has an X
and Y intercept. The x-intercept is (21.6 repeating, O) while the y-intercept is (0,6.5).
Next we find the Domain and Range. Since this equation gives us a line that will extend in both directions
infinitely the domain and range will simply be D=(-0,00) and R=(-0,00), or meaning all real numbers.
Next we determine if the equation is a function. To do so we can use the Vertical line test. Which is If a vertical
line crosses the Relation on the graph only once in all locations, the Relation is a Function. Also keep in mind
that all lines except vertical lines are Functions and closed shapes are not, but parabolic and exponential curves
are. So since my first equation is a straight line with a slope of -03 it passes the Vertical line test and is
a Function.
For my 2nd equation we have f(x)=3×2, so once again lets start of with plotting 5 points.
XY
-2 | 12
-13
00
13
2 | 12
For the second step we must determine if we have an X and Y intercept and what it is. Once again you by
looking at the points I’ve chosen you can see we have the convience of having our x-intercept also be are
y-intercept (0,0).
Now we find the Domain and Range for our equation. First the domain or as I personally like to think of it the
back and forth or right to left of the equation. As we control the x variable we can plug any number into the
equation to give us a y. So the Domain is simply (-0,00) going forever back and forth or right to left.
Now the Range for the range I think of it as the up and down. So for the range the lowest we can go is O and if
we use 0 as the x variable it will give us a y of O, however when we plug in a negative number it will always spit
out a positive y same goes for the positive side of the spectrum. For our range then we are left with (0,00) or
the lowest we will go is O but we will go on forever in the upwards direction.
Next we determine if the equation is a Function. To do so we will use the Vertical line test once again. Doing so
we can see that the equation is indeed a function as a vertical line will not touch the equation on the Yaxis
more than once for each given point on the X axis.
Unread
XY
-6 15
-56
-43
-36
-215
Now we have to rewrite the equation with the transformed points. To do so I will start by writing it in standard
form for a parabola y=a(x-h)2+k. To start we use our new vertex to substitute for hand k which will give us.
y=a(x+4)2+3 Now we choose any point on the parabola except the vertex so let’s go with (-3,6) to substitute for
x and y.
6=a(-3+4)2+3 Now all that remains is to solve for a which will give us a = 3. Then we just rewrite the equation.
y=3(x+4)2+3 which is the equation for our new parabola after the transformation.