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To show how to solve the equation, create two functions and find their intersection points. What two functions can be used to s...Asked by s
To show how to solve the equation, create two functions and find their intersection points. What two functions can be used to solve the following equation, and what is their solution set?
x2+4x−1=2x+2
𝑥
2
+
4
𝑥
−
1
=
2
𝑥
+
2
(1 point)
Responses
f(x)=x2+4x
𝑓
(
𝑥
)
=
𝑥
2
+
4
𝑥
, g(x)=2x
𝑔
(
𝑥
)
=
2
𝑥
, {−3,1}
{
−
3
,
1
}
f left parenthesis x right parenthesis equals x squared plus 4 x , g left parenthesis x right parenthesis equals 2 x , left brace negative 3 comma 1 right brace
f(x)=x2+4x−1
𝑓
(
𝑥
)
=
𝑥
2
+
4
𝑥
−
1
, g(x)=2x+2
𝑔
(
𝑥
)
=
2
𝑥
+
2
, {−3,−4,1,4}
{
−
3
,
−
4
,
1
,
4
}
f left parenthesis x right parenthesis equals x squared plus 4 x minus 1 , g left parenthesis x right parenthesis equals 2 x plus 2 , left brace negative 3 comma negative 4 comma 1 comma 4 right brace
f(x)=x2+4x−1
𝑓
(
𝑥
)
=
𝑥
2
+
4
𝑥
−
1
, g(x)=2x+2
𝑔
(
𝑥
)
=
2
𝑥
+
2
, {−1,3}
{
−
1
,
3
}
f left parenthesis x right parenthesis equals x squared plus 4 x minus 1 , g left parenthesis x right parenthesis equals 2 x plus 2 , left brace negative 1 comma 3 right brace
f(x)=x2+4x−1
𝑓
(
𝑥
)
=
𝑥
2
+
4
𝑥
−
1
, g(x)=2
x2+4x−1=2x+2
𝑥
2
+
4
𝑥
−
1
=
2
𝑥
+
2
(1 point)
Responses
f(x)=x2+4x
𝑓
(
𝑥
)
=
𝑥
2
+
4
𝑥
, g(x)=2x
𝑔
(
𝑥
)
=
2
𝑥
, {−3,1}
{
−
3
,
1
}
f left parenthesis x right parenthesis equals x squared plus 4 x , g left parenthesis x right parenthesis equals 2 x , left brace negative 3 comma 1 right brace
f(x)=x2+4x−1
𝑓
(
𝑥
)
=
𝑥
2
+
4
𝑥
−
1
, g(x)=2x+2
𝑔
(
𝑥
)
=
2
𝑥
+
2
, {−3,−4,1,4}
{
−
3
,
−
4
,
1
,
4
}
f left parenthesis x right parenthesis equals x squared plus 4 x minus 1 , g left parenthesis x right parenthesis equals 2 x plus 2 , left brace negative 3 comma negative 4 comma 1 comma 4 right brace
f(x)=x2+4x−1
𝑓
(
𝑥
)
=
𝑥
2
+
4
𝑥
−
1
, g(x)=2x+2
𝑔
(
𝑥
)
=
2
𝑥
+
2
, {−1,3}
{
−
1
,
3
}
f left parenthesis x right parenthesis equals x squared plus 4 x minus 1 , g left parenthesis x right parenthesis equals 2 x plus 2 , left brace negative 1 comma 3 right brace
f(x)=x2+4x−1
𝑓
(
𝑥
)
=
𝑥
2
+
4
𝑥
−
1
, g(x)=2
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve the equation \( x^2 + 4x - 1 = 2x + 2 \), we can set up two functions based on the left-hand side and the right-hand side of the equation:
1. Let \( f(x) = x^2 + 4x - 1 \)
2. Let \( g(x) = 2x + 2 \)
To find the solution set, we need to determine the intersection points of \( f(x) \) and \( g(x) \). This involves solving the equation:
\[ f(x) = g(x) \]
\[ x^2 + 4x - 1 = 2x + 2 \]
Next, we rearrange the equation by moving all terms to one side:
\[ x^2 + 4x - 1 - 2x - 2 = 0 \]
\[ x^2 + 2x - 3 = 0 \]
Now we can factor the quadratic equation:
\[ (x + 3)(x - 1) = 0 \]
Setting each factor equal to zero gives the possible solutions:
1. \( x + 3 = 0 \) ⇒ \( x = -3 \)
2. \( x - 1 = 0 \) ⇒ \( x = 1 \)
Thus, the solution set of the original equation \( x^2 + 4x - 1 = 2x + 2 \) is:
\[ \{ -3, 1 \} \]
### Summary:
- Functions:
- \( f(x) = x^2 + 4x - 1 \)
- \( g(x) = 2x + 2 \)
- Solution set:
- \( \{ -3, 1 \} \)
So the correct response is:
**\( f(x) = x^2 + 4x - 1 \), \( g(x) = 2x + 2 \), \( \{ -3, 1 \} \)**.
1. Let \( f(x) = x^2 + 4x - 1 \)
2. Let \( g(x) = 2x + 2 \)
To find the solution set, we need to determine the intersection points of \( f(x) \) and \( g(x) \). This involves solving the equation:
\[ f(x) = g(x) \]
\[ x^2 + 4x - 1 = 2x + 2 \]
Next, we rearrange the equation by moving all terms to one side:
\[ x^2 + 4x - 1 - 2x - 2 = 0 \]
\[ x^2 + 2x - 3 = 0 \]
Now we can factor the quadratic equation:
\[ (x + 3)(x - 1) = 0 \]
Setting each factor equal to zero gives the possible solutions:
1. \( x + 3 = 0 \) ⇒ \( x = -3 \)
2. \( x - 1 = 0 \) ⇒ \( x = 1 \)
Thus, the solution set of the original equation \( x^2 + 4x - 1 = 2x + 2 \) is:
\[ \{ -3, 1 \} \]
### Summary:
- Functions:
- \( f(x) = x^2 + 4x - 1 \)
- \( g(x) = 2x + 2 \)
- Solution set:
- \( \{ -3, 1 \} \)
So the correct response is:
**\( f(x) = x^2 + 4x - 1 \), \( g(x) = 2x + 2 \), \( \{ -3, 1 \} \)**.
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