Tag Archive: graphing


exponential functions

When graphing exponential functions, there are three things that you should remember.

  1. All exponential functions are CURVED.
  2. As the base increases, the graph shifts to the left of the y-axis.
  3. If you have a negative exponent, you must rewrite your base as a fraction.

ex. (1/2)^-2  would be rewritten as 1/1/2^2 = 4

**Remember, exponential functions are never straight lines. If your graph resembles a straight line, it is a linear function.

 

Section 3: Graphing Inequalities

Graph the following system of inequalities

2y<-x-2

Test point (0, 0)

False for the first statement; True for the second statement.

\

 

y=-3x+2 Click here to edit graph
y=3x-4

solution (1, -1)

There are two ways to graph a system of equations. If the equations are presented
in y-intercept form, then you would plot your first point at the y-intercept. Then,
you would use the slope (mx) to plot your remaining points. The solution will always be the location in which the two lines intersect.
Secondly, you can use the substitution method to find your x and y value for each equation. To prove that your solution is correct, graph each of the lines by placing them in y-intercept form. If the solution that you receive after you create your graph is the same as the solution you developed with using the substitution method, then your answer is correct.
ex.     x + y = 4
2x  – y = 5
Step 1: put the first equation into y-intercept form (y = -x + 4)
Step 2: substitute the new equation into the second-given equation (2x – (-x + 4) = 5)
Step 3: combine like terms (2x + x + 4 =5) –> (3x + 4 = 5) —> (3x – 4+4 = 5+4) —>
(3x/3 = 9/3) —>x = 3
Step 4: substitute the new equation into the first-given equation (3+ y = 4)
Step 5: subtract 1/3 from both sides of the equation (3-3 + y = 4 -3)
Step 6: solve (y = 4 – 3) —> y = 1
Step 7: write your solution ( 3, 1)
Step 8: make a graph to check your solution ( 3+1=4 ), ( 2(3) – 1 = 5) CHECK!

Click here to edit graph

Solve each system of equations by using the graphing or substitution method.

1. y= x + 3          x – 2y = 0

substitute your first equation for the y-value in the second equation.

x – 2(x + 3) = 0 —-> x- 2x – 6 = 0 —–> -x – 6 = 0 —–> -x -6 + 6 = 0 + 6 —-> -x/-1 = 6/-1 —-> x = -6

substitute your x-value into the first equation.

y= -6 + 3 —-> y = -3

write your solution

Solution (-6, -3)

x – 13 = 2y + x – 17          FIRST EQUATION

3x + y= 29                           SECOND EQUATION

Step 1: Take the 2nd equation and put it into y-intercept form.

3x – 3x + y = 29 – 3x——-> y  =-3x + 29

Step 2: Substitute the above equation for y in the 1st equation.

x – 13 = 2(-3x + 29) + x – 17 ——>     x- 13 = -6x + 58 + x -17 ——>  x – 13 = -5x + 41  ——> x + 5x – 13 = -5x + 5x + 41

6x – 13 = 41 —–> 6x = 54 —-> 6x/6 = 54/6 —–> x = 9

Step 3: Substitute your x-value into the second equation.

3(9) + y = 29 —–> 27 + y = 29 —–> 27- 27 + y = 29 – 27 ——> y = 2

Step 4: Write your solution.

Solution (9, 2)

Follow

Get every new post delivered to your Inbox.