Question

en Vision Algebra 2
1-1 Reteach to Build Understanding
Key Features of Functions
Linear, quadratic, and absolute-value functions can be graphed using key features
that the equation identifies. Key features can be labeled using either set builder
notation or interval notation. Interval notation represents a set of real numbers by
a pair of values. Parentheses are used when the points are not included. Brackets
are used for when a boundary point is included.
1. Use the function f(x) = x2 – 16 and identify the key features that are listed in
the table for the graph. Label the graph with the key features.
Key Feature
Domain
Range
Minimum Value
Function is Function is
x-intercept
decreasing increasing
10
y-intercept
LEAD) 14.0VA
Function is Positive
-2 O
Function is Negative
TO
Interval Increasing
Domain:
Interval Decreasing
Range:
-20160-16)

Answer 1

Answer:

Domain : $(-\infty,\infty)$, Range : $(-16,\infty)$, Minimum Value = -16

Function is a quadratic function. x-intercepts are -4 and 4.

The function decreasing before vertex = $(-\infty,0)$

The function increasing after vertex = $(0,\infty)$

y-intercept is -16.

Function is Positive : $(-\infty,-4)\cup (4,\infty)$, Function is negative : $(-4,4)$

Interval Increasing  Domain: $(0,\infty)$, Interval Decreasing  Range: $(-\infty,-16)$

Step-by-step explanation:

The given function is

$f(x)=x^2-16$    ...(1)

It is an upward parabola because coefficient of $x^2$ is positive.

The vertex form of a parabola is

$f(x)=a(x-h)^2+k$       ...(2)

where, a is constant and (h,k) is vertex.

From (1) and (2), we get

$a=1,h=0,k=-16$

So, vertex of the parabola is (0,-16).

Key features are:

The given function is defined for all values of x. So,

Domain : $(-\infty,\infty)$

The value of a function which is an upward parabola cannot be less than the y-coordinate of its vertex.

Range : $(-16,\infty)$

Minimum Value = y-coordinate of vertex = -16

Function is a quadratic function.

For f(x)=0,

$0=x^2-16$

$16=x^2$

$x=\pm 4$

So, x-intercepts are -4 and 4.

The function decreasing before vertex = $(-\infty,0)$

The function increasing after vertex = $(0,\infty)$

For x=0,

$f(0)=(0)^2-16=-16$

So, y-intercept is -16.

Function is Positive : $(-\infty,-4)\cup (4,\infty)$

Function is negative : $(-4,4)$

Interval Increasing  Domain: $(0,\infty)$

Interval Decreasing  Range: $(-\infty,-16)$