Answer :
Absolutely, let's delve into the definitions of the functions [tex]\( q(x) \)[/tex] and [tex]\( r(x) \)[/tex] and understand what they represent and how they can be utilized.
### Step-by-Step Solution:
First, let's write down the given definitions of the functions:
1. Function [tex]\( q(x) \)[/tex]:
[tex]\[ q(x) = x^2 + 6 \][/tex]
This is a quadratic function where:
- The term [tex]\( x^2 \)[/tex] represents a parabola that opens upwards.
- The constant [tex]\( 6 \)[/tex] shifts the entire parabola up by 6 units on the y-axis.
2. Function [tex]\( r(x) \)[/tex]:
[tex]\[ r(x) = \sqrt{x + 9} \][/tex]
This is a square root function where:
- The term [tex]\( x + 9 \)[/tex] means the square root is taken of the input shifted to the left by 9 units.
- The square root function is defined for [tex]\( x + 9 \geq 0 \)[/tex], implying [tex]\( x \geq -9 \)[/tex].
### Analysis of Functions:
1. Domain of [tex]\( q(x) \)[/tex]:
- Since [tex]\( q(x) = x^2 + 6 \)[/tex] involves a polynomial (squared term), [tex]\( x \)[/tex] can take any real number value.
- Therefore, the domain of [tex]\( q(x) \)[/tex] is all real numbers: [tex]\( (-\infty, \infty) \)[/tex].
2. Domain of [tex]\( r(x) \)[/tex]:
- Since [tex]\( r(x) = \sqrt{x + 9} \)[/tex], the expression inside the square root must be non-negative.
- Therefore, [tex]\( x + 9 \geq 0 \)[/tex], which simplifies to [tex]\( x \geq -9 \)[/tex].
- Hence, the domain of [tex]\( r(x) \)[/tex] is [tex]\( [-9, \infty) \)[/tex].
### Graphical Interpretation:
- The graph of [tex]\( q(x) = x^2 + 6 \)[/tex] is a parabola opening upwards with its vertex at [tex]\( (0, 6) \)[/tex].
- The graph of [tex]\( r(x) = \sqrt{x + 9} \)[/tex] is a square root function starting from the point [tex]\( (-9, 0) \)[/tex] and increasing slowly as [tex]\( x \)[/tex] increases.
### Value Calculation Examples:
1. Evaluate [tex]\( q(2) \)[/tex]:
[tex]\[ q(2) = 2^2 + 6 = 4 + 6 = 10 \][/tex]
2. Evaluate [tex]\( r(0) \)[/tex]:
\begin{align}
r(0) &= \sqrt{0 + 9} \\
&= \sqrt{9} \\
&= 3
\end{align}
### Understanding Functional Composition (If Required):
If you need to compose these functions, say [tex]\( q(r(x)) \)[/tex] or [tex]\( r(q(x)) \)[/tex], be aware of the domain restrictions and the resultant expressions.
- For [tex]\( q(r(x)) \)[/tex]:
\begin{align}
\text{First find } r(x): \, r(x) &= \sqrt{x + 9} \\
\text{Then find } q(r(x)): \, q(r(x)) &= q(\sqrt{x + 9}) \\
&= (\sqrt{x + 9})^2 + 6 \\
&= x + 9 + 6 \\
&= x + 15.
\end{align}
- For [tex]\( r(q(x)) \)[/tex]:
\begin{align}
\text{First find } q(x): \, q(x) &= x^2 + 6 \\
\text{Then find } r(q(x)): \, r(q(x)) &= r(x^2 + 6) \\
&= \sqrt{(x^2 + 6) + 9} \\
&= \sqrt{x^2 + 15}.
\end{align}
In summary:
- [tex]\( q(x) = x^2 + 6 \)[/tex] is a quadratic function valid for all real numbers.
- [tex]\( r(x) = \sqrt{x + 9} \)[/tex] is a square root function valid for [tex]\( x \geq -9 \)[/tex].
These functions can be analyzed, evaluated, and composed with attention to their domain constraints. Keep these details in mind when working with such functions in calculus or algebra problems.
### Step-by-Step Solution:
First, let's write down the given definitions of the functions:
1. Function [tex]\( q(x) \)[/tex]:
[tex]\[ q(x) = x^2 + 6 \][/tex]
This is a quadratic function where:
- The term [tex]\( x^2 \)[/tex] represents a parabola that opens upwards.
- The constant [tex]\( 6 \)[/tex] shifts the entire parabola up by 6 units on the y-axis.
2. Function [tex]\( r(x) \)[/tex]:
[tex]\[ r(x) = \sqrt{x + 9} \][/tex]
This is a square root function where:
- The term [tex]\( x + 9 \)[/tex] means the square root is taken of the input shifted to the left by 9 units.
- The square root function is defined for [tex]\( x + 9 \geq 0 \)[/tex], implying [tex]\( x \geq -9 \)[/tex].
### Analysis of Functions:
1. Domain of [tex]\( q(x) \)[/tex]:
- Since [tex]\( q(x) = x^2 + 6 \)[/tex] involves a polynomial (squared term), [tex]\( x \)[/tex] can take any real number value.
- Therefore, the domain of [tex]\( q(x) \)[/tex] is all real numbers: [tex]\( (-\infty, \infty) \)[/tex].
2. Domain of [tex]\( r(x) \)[/tex]:
- Since [tex]\( r(x) = \sqrt{x + 9} \)[/tex], the expression inside the square root must be non-negative.
- Therefore, [tex]\( x + 9 \geq 0 \)[/tex], which simplifies to [tex]\( x \geq -9 \)[/tex].
- Hence, the domain of [tex]\( r(x) \)[/tex] is [tex]\( [-9, \infty) \)[/tex].
### Graphical Interpretation:
- The graph of [tex]\( q(x) = x^2 + 6 \)[/tex] is a parabola opening upwards with its vertex at [tex]\( (0, 6) \)[/tex].
- The graph of [tex]\( r(x) = \sqrt{x + 9} \)[/tex] is a square root function starting from the point [tex]\( (-9, 0) \)[/tex] and increasing slowly as [tex]\( x \)[/tex] increases.
### Value Calculation Examples:
1. Evaluate [tex]\( q(2) \)[/tex]:
[tex]\[ q(2) = 2^2 + 6 = 4 + 6 = 10 \][/tex]
2. Evaluate [tex]\( r(0) \)[/tex]:
\begin{align}
r(0) &= \sqrt{0 + 9} \\
&= \sqrt{9} \\
&= 3
\end{align}
### Understanding Functional Composition (If Required):
If you need to compose these functions, say [tex]\( q(r(x)) \)[/tex] or [tex]\( r(q(x)) \)[/tex], be aware of the domain restrictions and the resultant expressions.
- For [tex]\( q(r(x)) \)[/tex]:
\begin{align}
\text{First find } r(x): \, r(x) &= \sqrt{x + 9} \\
\text{Then find } q(r(x)): \, q(r(x)) &= q(\sqrt{x + 9}) \\
&= (\sqrt{x + 9})^2 + 6 \\
&= x + 9 + 6 \\
&= x + 15.
\end{align}
- For [tex]\( r(q(x)) \)[/tex]:
\begin{align}
\text{First find } q(x): \, q(x) &= x^2 + 6 \\
\text{Then find } r(q(x)): \, r(q(x)) &= r(x^2 + 6) \\
&= \sqrt{(x^2 + 6) + 9} \\
&= \sqrt{x^2 + 15}.
\end{align}
In summary:
- [tex]\( q(x) = x^2 + 6 \)[/tex] is a quadratic function valid for all real numbers.
- [tex]\( r(x) = \sqrt{x + 9} \)[/tex] is a square root function valid for [tex]\( x \geq -9 \)[/tex].
These functions can be analyzed, evaluated, and composed with attention to their domain constraints. Keep these details in mind when working with such functions in calculus or algebra problems.
