Unit+1+Journals

1.2:

In step 1 the 10 is not outside of the brackets but it is added in step 2 which means you have to distribute it. First you distribute the 10 to the 1/2x which basically means divide 10 by 2, giving you 5x. Distributing the 10 to the 4 just means multiply them so it is 5x+40. To distribute the 10 to 3/5, just multiply 10 by 3 which equals 30 and divide by 5, giving you 6x. Lastly in step 2 you multiply 10 by -2 which is -20. That is why step 3 is 5x+40=6x-20. What you are trying to do next in step 3 to get to step 4 is getting x by itself, meaning you have to first subtract one of the x values from both sides. That means you subtract 6x from both sides giving you the -1x+40=-20 in step 4. Next you have to subtract 40 from both sides to get x by itself so -20-40 is -60. This is how you get -1x=-60 in step 5. To get x=60 in step 6, divide both sides of the equation by -1 and that gives you x=60. Now since you have the value of x you have to plug it in to the original equation which is what step 7 is. Step 7 is 1/2(60)+4=3/5(60)-2. Divide 60 in half which is 30 and multiply 60 by the numerator 3 and divide by the denominator 5 giving you 36. So what you have in step 8 is 30+4=36-2. Do the simple addition and subtraction and you will get the 34=34 that is in step 9.

1.3:

The type of problem that is easiest for me to solve is obviously the simple algebraic equations that doesn't have that many steps. Function notation isn't that hard but I have to be careful with the graphic aspect of it so I don't incorrectly plot the points. Now that I know that a point can have two x or y values plotted on the graph, I have to look carefully at the graph because if you don't name both points the answer is wrong. Also, we just learned about domains and I have to remember when to use the brackets and when to use the parentheses.

-Use brackets when the point plotted is a closed circle -Use parentheses when it's infinity or an open circle on the graph -In function notation, f(x)=y -If x is given then you plug it into the equation and solve for y

1.4:

~Linear Function- An equation with variables that you have to solve for ~Relation- How two items are connected to each other ~Domain- All of the x values ~Range- All of the y values ~Increasing- Getting larger ~Decreasing- Getting smaller ~Slope- The rise over the run that indicates how steep a line is that is represented by m ~Intercept- The point in which a line intersects with the x or y axis ~Degree- The highest power of a variable

1.5:

A __relation__ is an input value and its output value. A __linear function__ is a relation in the form of y=mx+b.

An example of a function is (6,3) and (7,4). An example that is not a function is (8,5) and (9,5). It is not a function because both points have the same y value.



f(x)=2x+3

The slope is 2 and the y-intercept is 3. (2,7) (3,9) and (-3,-3) are all on the line.

1.6:

The equation that matches up with Graph A is number 8, (f)x=-1/3x+4. The equation that matches up with Graph B is number 2, (f)x=2x+4. The equation that matches up with Graph C is number 6, (f)x=-1/2x-4. The equation that matches up with Graph D is number 7, (f)x=1/3x+4. The first thing I did was look at the graph. I located where the point is when x is 0 which is the y-intercept. If you know the y-intercept then you can narrow down which equation matches up with it. Then I see how many points you had to go up or down, left or right, and matched that with the equation that had the corresponding slope.

1.7:

The equations for Graph A are number 4 [(f)x=-2x+3] and letter e: 2x+3y=-3. The equations for Graph B are number 1 [(f)x=-2/3x+1] and letter e: 2x+3y=-3. The equations for Graph C are number 5 [f(x)=3x+2] and letter e: 2x+3y=-3. The equations for Graph D are number 3[(f)x=2x+3] and letter b: 2x+3y=3. First what I did was look at the y-intercept of the graphs and found the equation with the corresponding y-intercept. I also looked at the slope of the graph to see if it was negative or positive and then I looked at the equations with the negative or positive slopes. Once I found the matching equation in slope-intercept form, I found the corresponding equation in standard form.

1.8: The graph shows Anne's speed in miles per hour as she drove from home to school. At one minute of driving, Anne was at 30 mph. She continued to drive at 30 mph up until 4 minutes. From 4.25 to 5 minutes, she wasn't moving at all. Then from 6 minutes to 8 minutes, Anne drove at 40 mph. She slowed down and drove 20 mph from 9 minutes to 11.5 minutes, and then stopped driving at 12 minutes. (in minutes) || Speed (in miles per hour) || 1. Anne is driving the fastest from the 6-8 minute mark. I know this because on the graph the line is above all the other points.
 * Time
 * 0 || 0 ||
 * 1 || 30 ||
 * 4 || 30 ||
 * 4.25 || 0 ||
 * 5 || 0 ||
 * 6 || 40 ||
 * 8 || 40 ||
 * 9 || 20 ||
 * 11.5 || 20 ||
 * 12 || 0 ||

2. Anne is stopped at 0 minute mark, from 4.25-5 minutes, and at the 12 minute mark. I know this because her speed is 0 mph at those times.

3. Anne's speed decreases at 4 minutes, 8 minutes, and 11.5 minutes. These points on the graph are where the line starts to decrease.

4. Anne's speed at 7 minutes is 40 mph.

5. Anne is approximately driving 35 mph at the 5.75 and 8.25 minute mark.

1.9:

I observed that in the f(x)=1/2x table, f(x) is the x divided by two. I also noticed that in both tables f(x) increases by .5 each time.