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?

Question

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

Answer 1

To solve the equation \( x^2 + 4x - 1 = 2x + 2 \), we can rearrange it to form a standard equation by moving all terms to one side:

\[ x^2 + 4x - 1 - 2x - 2 = 0 \] \[ x^2 + 2x - 3 = 0 \]

Now we can create two functions based on the original equation:

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

To find the solutions to the equation, we look for the intersection points of the two functions \( f(x) \) and \( g(x) \). This means we need to solve the equation \( f(x) = g(x) \).

Alternatively, we can use the quadratic equation \( x^2 + 2x - 3 = 0 \) to find the roots:

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1 \), \( b = 2 \), and \( c = -3 \):

\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \] \[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} \] \[ x = \frac{-2 \pm \sqrt{16}}{2} \] \[ x = \frac{-2 \pm 4}{2} \]

Calculating the two possible solutions:

1. \( x = \frac{2}{2} = 1 \)
2. \( x = \frac{-6}{2} = -3 \)

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

So the correct response is:

f(x)=x²+4x−1, g(x)=2x+2, {-3,1}