Question

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The Usefulness of Graphs Quick Check
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Question
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

(1 point)
Responses

f(x)=x2+4x−1
, g(x)=2x+2
, {−3,1}
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 1 right brace

f(x)=x2+4x
, g(x)=2x
, {−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
, g(x)=2x+2
, {−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
, g(x)=2x+2
, {−3,−4,1,4}

Answer 1

To solve the equation \( x^2 + 4x - 1 = 2x + 2 \) by creating two functions, we can define:

1. \( f(x) = x^2 + 4x - 1 \)
2. \( g(x) = 2x + 2 \)

Next, we need to find the intersection points of these two functions, which represent the solutions to the equation. This can be done by setting \( f(x) \) equal to \( g(x) \):

\[
x^2 + 4x - 1 = 2x + 2
\]

Rearranging the equation gives:

\[
x^2 + 4x - 2x - 1 - 2 = 0
\]

This simplifies to:

\[
x^2 + 2x - 3 = 0
\]

Now we can factor the quadratic:

\[
(x + 3)(x - 1) = 0
\]

Thus, the solutions are:

\[
x = -3 \quad \text{and} \quad x = 1
\]

The solution set is \( \{-3, 1\} \).

So, the correct response is:

\( f(x) = x^2 + 4x - 1, \, g(x) = 2x + 2, \, \{-3, 1\} \)