into this equation. There are no x-values that
will generate an undefined or imaginary y
value. However, it is impossible to generate a
y-value that is less than –3. Any x-value greater
than or less than 3 will generate a yvalue that
is greater than –3. Therefore, the range of the
equation y= |x| –3 is all real numbers greater
than or equal to –3. The domain and range of
y= |x| –3 are not the same. The equation of
the graph in diagram Eis y= (x– 3)^2 + 1.
With this equation as well, any real number
can be substituted for x—there are no x-values
that will generate an undefined or imaginary
y-value. However, it is impossible to generate a
y-value that is less than 1. Any x-value greater
than or less than 3 will generate a y-value that
is greater than 1. Therefore, the range of the
equation y= (x– 3)^2 + 1 is all real numbers
greater than or equal to 1. The domain and
range ofy= (x– 3)^2 + 1 are not the same.
- e.The graph of the equation in diagram Ais
not a function. It fails the vertical line test for
all x-values where –2x2. The equation of
the graph in diagram Bis y= |x| – 3. Any x-
value greater than or less than 3 will generate a
y-value that is greater than –3; no values less
than –3 can be generated by this equation.
Therefore, the range of the equation y= |x| – 3
is all real numbers greater than or equal to –3.
The equation of the graph in diagram Cis y=
x. Since the square roots of negative num-
bers are imaginary, the range of this equation
is all real numbers greater than or equal to 0.
The equation of the graph in diagramDis y=
^1 x. Dividing 1 by a real number (excluding 0)
will yield real numbers, excluding 0. There is
no value for xthat can make y= 0. Therefore,
the range of the equation y= ^1 xis all real numbers
excluding 0. The equation of the graph in
diagram Eis y= (x– 3)^2 + 1. Any x-value
greater than or less than 3 will generate a y
value that is greater than 1; no values less than
1 can be generated by this equation. Therefore,
the range of the equation y= (x– 3)^2 + 1 is all
real numbers greater than or equal to 1. Of the
four equations that are functions, the equation
y= (x– 3)^2 + 1 (E), has the smallest range
(fewest elements), since the set of real num-
bers that are greater than or equal to 1 is
smaller than the set of all real numbers greater
than or equal to –3 (B), smaller than the set of
all real numbers greater than or equal to 0 (C),
and smaller than the set of all real numbers
excluding 0 (D).
- b.Substitute the expression 2y– 1 for every
occurrence ofxin the definition of the func-
tion f(x), and then simplify:
f(2y– 1) =
(2y– 1)^2 + 3(2y– 1) – 2 =
4 y^2 – 4y+ 1 + 6y– 3 – 2 =
4 y^2 + 2y– 4
- c.Simplifying f(x+ h) requires we substitute
the expression x+ h for every occurrence ofx
in the definition of the function f(x), and then
simplify:
f(x+ h) =
–((x+ h) –1)^2 + 3 =
–[(x+ h)^2 – 2(x+ h)+ 1] + 3 =
–[x^2 + 2hx+ h^2 – 2x– 2h+ 1] + 3 =
- x^2 – 2hx– h^2 + 2x+ 2h– 1 + 3 =
- x^2 – 2hx– h^2 + 2x+ 2h+ 2
Next, in anticipation of simplifying f(x+ h) – f(x),
we simplify the expression for f(x) = –(x– 1)^2 + 3
in order to facilitate combining like terms:
ANSWERS & EXPLANATIONS–
