With the function that you entered of 2x - 5, plot points, determine the intercepts, domain, range
Since you did not specify a qualifying variable or function notation in your expression, we will assume y
y = 2x - 5
Determine function type:
Since we have a variable with no exponents:
this is a
linear function
Since this is a linear function
it is a direct variation equation. The constant of proportionality is 2
Now Plot points from 10 to -10
x | Plug in x | ƒ(x) = 2x - 5 | Ordered Pair | -10 | 2(-10)-5 | -25 | (-10, -25) |
-9 | 2(-9)-5 | -23 | (-9, -23) |
-8 | 2(-8)-5 | -21 | (-8, -21) |
-7 | 2(-7)-5 | -19 | (-7, -19) |
-6 | 2(-6)-5 | -17 | (-6, -17) |
-5 | 2(-5)-5 | -15 | (-5, -15) |
-4 | 2(-4)-5 | -13 | (-4, -13) |
-3 | 2(-3)-5 | -11 | (-3, -11) |
-2 | 2(-2)-5 | -9 | (-2, -9) |
-1 | 2(-1)-5 | -7 | (-1, -7) |
0 | 2(0)-5 | -5 | (0, -5) |
1 | 2(1)-5 | -3 | (1, -3) |
2 | 2(2)-5 | -1 | (2, -1) |
3 | 2(3)-5 | 1 | (3, 1) |
4 | 2(4)-5 | 3 | (4, 3) |
5 | 2(5)-5 | 5 | (5, 5) |
6 | 2(6)-5 | 7 | (6, 7) |
7 | 2(7)-5 | 9 | (7, 9) |
8 | 2(8)-5 | 11 | (8, 11) |
9 | 2(9)-5 | 13 | (9, 13) |
10 | 2(10)-5 | 15 | (10, 15) |
Determine the y-intercept:
The y-intercept is found when x is set to 0. From the grid above, our y-intercept is -5
Determine the x-intercept
The x-intercept is found when y is set to 0
The y-intercept is found when y is set to 0. From the grid above, our x-intercept is 0
Determine the domain of the function:
The domain represents all values of x that you can enter
The domain is (-∞, ∞) or All Real Numbers
Determine the range of the function:
The range is all the possible values of y or ƒ(x) that can exist
The range is (-∞, ∞) or All Real Numbers
(-10, -25)
(-9, -23)
(-8, -21)
(-7, -19)
(-6, -17)
(-5, -15)
(-4, -13)
(-3, -11)
(-2, -9)
(-1, -7)
(0, -5)
(1, -3)
(2, -1)
(3, 1)
(4, 3)
(5, 5)
(6, 7)
(7, 9)
(8, 11)
(9, 13)
(10, 15)