a) We'll use the slope- intercept form of the equation of
linear function:
y = mx + n, where m is the slope and n is
the y intercept of the line.
In the given case, when the
slope m = 2, we can write the equation:
y = 2x +
n
To identify several members of this family, we'll plug in
values for n:
n = 1 => y = 2x +
1
n = 2 => y = 2x + 2
n
= `sqrt(2)` => y = 2x + `sqrt(2)`
Therefore, the
number of members of the family of linear functions whose slope is m=2 is infinite,
because the number of real values of n is infinite, too.
b)
To find an equation of a linear function, respecting the given condition f(2) = 1, we'll
recall the form of a linear function:
f(x) = mx +
n
If x = 2 => f(2) = 2m +
n
If f(2) = 1 => 2m + n = 1 => m = (1 -
n)/2
We'll write the linear function in terms of x and
n:
f(x) = (1-n)x/2 +
n
Therefore, the number of members of the family of linear
function f(x) = (1-n)x/2 + n is infinite, because the values of n belong to the real set
of numbers.
c) To discover a member that belongs to both
families of linear functions, we'll have to impose the following constraint: the slopes
must be equal and the y intercepts must be also equal.
2 =
(1 - n)/2
4 = 1 - n
n = 1 -
4
n =
-3
Therefore, the member that belongs to both
families of linear functions is y = 2x - 3.
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