Question

complete the table of values for the functions $f(x) = (x + 2)^2 - 3$ and $g(x) = x + 1$. write your answers as whole numbers, decimals, or simplified fractions or mixed numbers. \

$$\begin{tabular}{|c|c|c|} \\hline \ owcolor{orange} $x$ & $f(x)$ & $g(x)$ \\\\ \\hline $-2$ & \\framebox{} & \\framebox{} \\\\ \\hline $-1$ & \\framebox{} & \\framebox{} \\\\ \\hline $0$ & \\framebox{} & \\framebox{} \\\\ \\hline $1$ & \\framebox{} & \\framebox{} \\\\ \\hline $2$ & \\framebox{} & \\framebox{} \\\\ \\hline \\end{tabular}$$

based on the values in the table, where does the equation $f(x) = g(x)$ have a solution? $x = -1$; between $x = -1$ and $x = 0$; $x = 0$; between $x = 0$ and $x = 1$

Explanation:

Part 1: Completing the table for \( f(x) = (x + 2)^2 - 3 \) and \( g(x) = x + 1 \)

For \( x = -2 \):

• Step 1: Calculate \( f(-2) \)

Substitute \( x = -2 \) into \( f(x) \):
\( f(-2) = (-2 + 2)^2 - 3 = (0)^2 - 3 = -3 \)

• Step 2: Calculate \( g(-2) \)

Substitute \( x = -2 \) into \( g(x) \):
\( g(-2) = -2 + 1 = -1 \)

For \( x = -1 \):

• Step 1: Calculate \( f(-1) \)

Substitute \( x = -1 \) into \( f(x) \):
\( f(-1) = (-1 + 2)^2 - 3 = (1)^2 - 3 = 1 - 3 = -2 \)

• Step 2: Calculate \( g(-1) \)

Substitute \( x = -1 \) into \( g(x) \):
\( g(-1) = -1 + 1 = 0 \)

For \( x = 0 \):

• Step 1: Calculate \( f(0) \)

Substitute \( x = 0 \) into \( f(x) \):
\( f(0) = (0 + 2)^2 - 3 = (2)^2 - 3 = 4 - 3 = 1 \)

• Step 2: Calculate \( g(0) \)

Substitute \( x = 0 \) into \( g(x) \):
\( g(0) = 0 + 1 = 1 \)

For \( x = 1 \):

• Step 1: Calculate \( f(1) \)

Substitute \( x = 1 \) into \( f(x) \):
\( f(1) = (1 + 2)^2 - 3 = (3)^2 - 3 = 9 - 3 = 6 \)

• Step 2: Calculate \( g(1) \)

Substitute \( x = 1 \) into \( g(x) \):
\( g(1) = 1 + 1 = 2 \)

For \( x = 2 \):

• Step 1: Calculate \( f(2) \)

Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = (2 + 2)^2 - 3 = (4)^2 - 3 = 16 - 3 = 13 \)

• Step 2: Calculate \( g(2) \)

Substitute \( x = 2 \) into \( g(x) \):
\( g(2) = 2 + 1 = 3 \)

The completed table is:

\( x \)	\( f(x) \)	\( g(x) \)
\( -1 \)	\( -2 \)	\( 0 \)
\( 0 \)	\( 1 \)	\( 1 \)
\( 1 \)	\( 6 \)	\( 2 \)
\( 2 \)	\( 13 \)	\( 3 \)

Part 2: Finding where \( f(x) = g(x) \)

We analyze the values of \( f(x) \) and \( g(x) \):

• At \( x = -1 \): \( f(-1) = -2 \), \( g(-1) = 0 \) (so \( f(x) < g(x) \))
• At \( x = 0 \): \( f(0) = 1 \), \( g(0) = 1 \) (so \( f(x) = g(x) \) here? Wait, no—wait, \( f(0) = 1 \) and \( g(0) = 1 \), so \( x = 0 \) is a solution? Wait, but let’s check again. Wait, \( f(0) = (0 + 2)^2 - 3 = 4 - 3 = 1 \), and \( g(0) = 0 + 1 = 1 \). So \( f(0) = g(0) \). But let’s check the options. Wait, the options are:

• \( x = -1 \)
• between \( x = -1 \) and \( x = 0 \)
• \( x = 0 \)
• between \( x = 0 \) and \( x = 1 \)

Wait, at \( x = 0 \), \( f(x) = g(x) = 1 \), so \( x = 0 \) is a solution. But let’s confirm:

Wait, when \( x = 0 \), \( f(0) = (0 + 2)^2 - 3 = 4 - 3 = 1 \), and \( g(0) = 0 + 1 = 1 \). So \( f(0) = g(0) \), so \( x = 0 \) is a solution.

Final Answers

Table Completion:

\( x \)	\( f(x) \)	\( g(x) \)
\( -1 \)	\( -2 \)	\( 0 \)
\( 0 \)	\( 1 \)	\( 1 \)
\( 1 \)	\( 6 \)	\( 2 \)
\( 2 \)	\( 13 \)	\( 3 \)

Solution to \( f(x) = g(x) \):

\( x = 0 \) (since \( f(0) = g(0) = 1 \))

(Note: If there was a miscalculation earlier, but based on the table, at \( x = 0 \), both functions equal 1, so \( x = 0 \) is the solution.)

Answer:

Part 1: Completing the table for \( f(x) = (x + 2)^2 - 3 \) and \( g(x) = x + 1 \)

For \( x = -2 \):

• Step 1: Calculate \( f(-2) \)

Substitute \( x = -2 \) into \( f(x) \):
\( f(-2) = (-2 + 2)^2 - 3 = (0)^2 - 3 = -3 \)

• Step 2: Calculate \( g(-2) \)

Substitute \( x = -2 \) into \( g(x) \):
\( g(-2) = -2 + 1 = -1 \)

For \( x = -1 \):

• Step 1: Calculate \( f(-1) \)

Substitute \( x = -1 \) into \( f(x) \):
\( f(-1) = (-1 + 2)^2 - 3 = (1)^2 - 3 = 1 - 3 = -2 \)

• Step 2: Calculate \( g(-1) \)

Substitute \( x = -1 \) into \( g(x) \):
\( g(-1) = -1 + 1 = 0 \)

For \( x = 0 \):

• Step 1: Calculate \( f(0) \)

Substitute \( x = 0 \) into \( f(x) \):
\( f(0) = (0 + 2)^2 - 3 = (2)^2 - 3 = 4 - 3 = 1 \)

• Step 2: Calculate \( g(0) \)

Substitute \( x = 0 \) into \( g(x) \):
\( g(0) = 0 + 1 = 1 \)

For \( x = 1 \):

• Step 1: Calculate \( f(1) \)

Substitute \( x = 1 \) into \( f(x) \):
\( f(1) = (1 + 2)^2 - 3 = (3)^2 - 3 = 9 - 3 = 6 \)

• Step 2: Calculate \( g(1) \)

Substitute \( x = 1 \) into \( g(x) \):
\( g(1) = 1 + 1 = 2 \)

For \( x = 2 \):

• Step 1: Calculate \( f(2) \)

Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = (2 + 2)^2 - 3 = (4)^2 - 3 = 16 - 3 = 13 \)

• Step 2: Calculate \( g(2) \)

Substitute \( x = 2 \) into \( g(x) \):
\( g(2) = 2 + 1 = 3 \)

The completed table is:

\( x \)	\( f(x) \)	\( g(x) \)
\( -1 \)	\( -2 \)	\( 0 \)
\( 0 \)	\( 1 \)	\( 1 \)
\( 1 \)	\( 6 \)	\( 2 \)
\( 2 \)	\( 13 \)	\( 3 \)

Part 2: Finding where \( f(x) = g(x) \)

We analyze the values of \( f(x) \) and \( g(x) \):

• At \( x = -1 \): \( f(-1) = -2 \), \( g(-1) = 0 \) (so \( f(x) < g(x) \))
• At \( x = 0 \): \( f(0) = 1 \), \( g(0) = 1 \) (so \( f(x) = g(x) \) here? Wait, no—wait, \( f(0) = 1 \) and \( g(0) = 1 \), so \( x = 0 \) is a solution? Wait, but let’s check again. Wait, \( f(0) = (0 + 2)^2 - 3 = 4 - 3 = 1 \), and \( g(0) = 0 + 1 = 1 \). So \( f(0) = g(0) \). But let’s check the options. Wait, the options are:

• \( x = -1 \)
• between \( x = -1 \) and \( x = 0 \)
• \( x = 0 \)
• between \( x = 0 \) and \( x = 1 \)

Wait, at \( x = 0 \), \( f(x) = g(x) = 1 \), so \( x = 0 \) is a solution. But let’s confirm:

Wait, when \( x = 0 \), \( f(0) = (0 + 2)^2 - 3 = 4 - 3 = 1 \), and \( g(0) = 0 + 1 = 1 \). So \( f(0) = g(0) \), so \( x = 0 \) is a solution.

Final Answers

Table Completion:

\( x \)	\( f(x) \)	\( g(x) \)
\( -1 \)	\( -2 \)	\( 0 \)
\( 0 \)	\( 1 \)	\( 1 \)
\( 1 \)	\( 6 \)	\( 2 \)
\( 2 \)	\( 13 \)	\( 3 \)

Solution to \( f(x) = g(x) \):

\( x = 0 \) (since \( f(0) = g(0) = 1 \))

(Note: If there was a miscalculation earlier, but based on the table, at \( x = 0 \), both functions equal 1, so \( x = 0 \) is the solution.)