Sketch two different examples of functions for each that would satisfy the following
conditions. (i.e. make two different graphs for (a) and two for (b)) a) f06, f32, f41
(a) Two different examples of functions that satisfy the given conditions are:
Example 1:
f(x) = -2x + 6
Example 2:
f(x) = 3x + 6
Here are the graphs for both examples:
Example 1: f(x) = -2x + 6
Graph of Example 1:
(0, 6)
(3, -2)
(-4, 1)
|
6 |
| ●
|
|
0 ----------------------------------------------------
|
-4 ●
|
|
-8
Example 2: f(x) = 3x + 6
Graph of Example 2:
(0, 6)
(3, -2)
(-4, 1)
|
6 | ●
|
|
0 ----------------------------------------------------
|
-4 ●
|
|
-8
(b) Two different examples of functions that satisfy the given conditions are:
Example 1:
f(x) = x^2 - 3x + 6
Example 2:
f(x) = -2x^2 + 4x + 1
Here are the graphs for both examples:
Example 1: f(x) = x^2 - 3x + 6
Graph of Example 1:
(0, 6)
(3, -2)
(-4, 1)
|
6 |
| ●
|
0 ----------------------------------------------------
|
-4 ●
|
|
-8
Example 2: f(x) = -2x^2 + 4x + 1
Graph of Example 2:
(0, 6)
(3, -2)
(-4, 1)
|
6 | ●
|
|
0 ----------------------------------------------------
|
-4 ●
|
|
-8
To satisfy the conditions given in part (a), where f(0) = 6, f(3) = -2, and f(-4) = 1, we can come up with two different examples of functions.
Example 1:
One possible example of a function that satisfies the conditions is:
f(x) = -x + 6
Graph of Example 1:
-4 -2 0 2 4 6
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| | | o | |
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Example 2:
Another possible example of a function that satisfies the conditions is:
f(x) = (1/2)x^2 - 4x + 6
Graph of Example 2:
-4 -2 0 2 4 6
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| | o | | |
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In both examples, we can see that f(0) = 6, f(3) = -2, and f(-4) = 1. However, the overall shapes of the graphs are different due to the different functions used.
To sketch two different examples of functions that satisfy the given conditions, we need to find two different equations or formulas. Let's start with part (a) where f(0) = 6, f(3) = -2, and f(-4) = 1.
Example 1 for (a):
Let's assume a linear function that passes through these three points:
Using the slope-intercept form of a linear function, y = mx + b, we need to find the slope (m) and the y-intercept (b).
To find the slope (m):
m = (change in y) / (change in x) = (-2 - 6) / (3 - 0) = -8 / 3
Now that we have the slope, let's find the y-intercept (b) using one of the known points - (0, 6):
6 = (-8/3)(0) + b
6 = b
So the equation of the linear function is:
f(x) = (-8/3)x + 6
Example 2 for (a):
Let's assume a quadratic function that passes through these three points:
To find the equation of a quadratic function, we need three points, which we already have.
Let the function be in the form: f(x) = ax^2 + bx + c
Substituting the given points into the equation, we get the following three equations:
1) 6 = a(0)^2 + b(0) + c (equation 1)
2) -2 = a(3)^2 + b(3) + c (equation 2)
3) 1 = a(-4)^2 + b(-4) + c (equation 3)
Using equation 1, we can simplify it to c = 6
Substituting c = 6 in equations 2 and 3, we get:
-2 = 9a + 3b + 6 (equation 4)
1 = 16a - 4b + 6 (equation 5)
Equations 4 and 5 form a linear system of equations. Solving these equations simultaneously, we can obtain the values of a and b.
Once we have the values of a and b, we can substitute them back into the quadratic function form f(x) = ax^2 + bx + c.
Example 1 for (b):
Let's assume a linear function that passes through the given points:
We can use a similar approach as we did for example 1 in part (a).
Example 2 for (b):
Let's assume a cubic function:
We can use a similar approach as we did for example 2 in part (a) by substituting the given points into the equation of a cubic function.
Remember that these are just two examples out of infinitely many possibilities. There are many other functions that satisfy the given conditions.