c Is the function defined for all x? Does the function increase or decrease indefinitely?
A
aDomain of f(x): All real numbers
Range of f(x): y ≥ 0
B
bPossible Solutions: g(x)≈ 2(x-5)^2+2 and h(x)≈ - 1/2(x-5)^2+2
C
cDomain of g(x): All real numbers
Range of g(x): y ≥ 3
Practice makes perfect
a The function f(x)=x^2 is defined for all x, so its domain is the set of all real numbers. Looking at the diagram, we see that the function's minimal value is y=0. Since we also see that the function increases indefinitely, we see that its range is y ≥ 0.
b From the diagram, we can determine a few things. First of all, two of the functions have a vertex at (6,3).
Since we know the vertex of the unknown functions, we can write them both in graphing form.
Graphing Form:& y=a(x-h)^2+ k
Vertex:& (h, k)
By substituting the known vertex in this equation, we can start writing the functions.
g(x)&=a(x-6)^2+ 3
h(x)&=a(x-6)^2+ 3
Next, we want to find the value of a. This determines if a curve opens upward or downward and if there is any vertical stretch or shrink.
Transformations of f(x)
Vertical Stretch or Shrink
Vertical stretch, a >1
y= af(x)
Vertical shrink, 0< a < 1
y= af(x)
Properties of f(x)
Direction of opening
upward, a > 0
y= af(x)
downward, a < 0
y= af(x)
Since f(x) opens upward, we know that it has a positive value of a. Conversely, because g(x) opens downward, the value of a is negative. It appears that f(x) is vertically stretched while g(x) is vertically shrunk. With this information, we can box in the value of a for our functions.
&g(x) ⇒ a >1
&h(x) ⇒ - 1 < a <0
We do not know exactly where in these intervals the value of a falls. This is where it becomes an approximation.
g(x)&≈ 2(x-6)^2+3
h(x)&≈ - 1/2(x-6)^2+3
c From the diagram we see that the function g(x) is defined for all x, so its domain is the set of all real numbers. We also see that the function's minimal value is y=3. Since the function increases indefinitely, we see that its range is y ≥ 3.
