Answer:
- a. linear: y = 42-2x
- b. non-linear: y = x(x +3)/2
Step-by-step explanation:
A function is linear if x-values are evenly spaced (all have the same difference) and y-values are evenly spaced (all have the same difference).
a) x-values have a difference of 6 -3 = 9 -6 = 12 - 9 = 3. y-values have a difference of 30 -36 = 24 -30 = 18 -24 = -6. Both these differences are constant, so the function is <em>linear</em>.
The ratio of y-differences to x-differences is -6/3 = -2, so that is the slope of the line. We can use the point-slope form to discover an equation in slope-intercept form.
Point-slope form of the equation for a line with slope m through point (h, k) can be written as ...
y = m(x -h) +k
Here, we have m = -2, and the first point is (h, k) = (3, 36). Then our line's equation can be written as ...
y = -2(x -3) +36
y = -2x +42 . . . . . . . eliminating parentheses
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b) The x-values in this table have a constant difference of ...
3 -1 = 5 -3 = 7 -5 = 2
The y-values have differences of 9 -2 = 7, 20 -9 = 11, 35 -20 = 15. These are not constant, so the relation is <em>non-linear</em>. These first differences have differences of ...
11 -7 = 15 -11 = 4 . . . . . second differences are constant
When second differences are constant, the relation can be described by a second degree polynomial. We can write some equations to discover what that polynomial is.
Generic form:
ax² +bx +c = y
Filling in three of the given points, we have three equations in a, b, c:
a·1² + b·1 +c = 2
a·3² +b·3 +c = 9
a·5² +b·5 +c = 20
Subtracting the first equation from the other two eliminates c and gives two equations in a and b:
a·(9 -1) +b(3 -1) = 9 -2 . . . . . . . . . 8a +2b = 7
a·(25 -1) + b·(5 -1) = 20 -2 . . . . . 24a +4b = 18
Subtracting twice the first of these equations from the second, we can eliminate b:
a(24 -2·8) = 18 -2·7
8a = 4
a = 1/2 . . . . . divide by 8
Substituting this into the first of the equations in a and b, we get:
8·(1/2) +2b = 7
2b = 3 . . . . . subtract 4
b = 3/2 . . . . divide by 2
Substituting for a and b in the first of our original equations, we find ...
(1/2)·1 + (3/2)·(1) +c = 2
2 +c = 2 . . . simplify
c = 0 . . . . . . subtract 2
So, the table in part b can be described by the quadratic equation ...
y = (1/2)x² + (3/2)x
