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

10−x2=x+4


(1 point)
Responses

g(x)=x+4
, f(x)=10−x2
, {1,6}
g left parenthesis x right parenthesis equals x plus 4 , f left parenthesis x right parenthesis equals 10 minus x squared , left brace 1 comma 6 right brace

g(x)=x+4
, f(x)=10+x2
, {−3,2}
g left parenthesis x right parenthesis equals x plus 4 , f left parenthesis x right parenthesis equals 10 plus x squared , left brace negative 3 comma 2 right brace

f(x)=10−x2
, g(x)=x+4
, {−3,2}
f left parenthesis x right parenthesis equals 10 minus x squared , g left parenthesis x right parenthesis equals x plus 4 , left brace negative 3 comma 2 right brace

f(x)=10+x2
, g(x)=x+4
, {−3,1}

Answer 1

To solve the equation \(10 - x^2 = x + 4\), we can create two functions and find their intersections.

Let:

• \(f(x) = 10 - x^2\)
• \(g(x) = x + 4\)

Now, we want to find the intersection points of these two functions. This means we need to solve the equation:

\[ 10 - x^2 = x + 4 \]

Rearranging the equation, we can express it as:

\[ 10 - x - 4 - x^2 = 0 \]

Simplifying this gives:

\[ -x^2 - x + 6 = 0 \]

To make it easier, we can multiply through by -1:

\[ x^2 + x - 6 = 0 \]

Next, we factor the quadratic equation:

\[ (x + 3)(x - 2) = 0 \]

Setting each factor to zero gives the solutions:

1. \(x + 3 = 0 \Rightarrow x = -3\)
2. \(x - 2 = 0 \Rightarrow x = 2\)

Thus, the solution set is:

\({-3, 2}\)

Based on the above analysis, the correct choice from the responses would be:

f(x) = 10 - x^2, g(x) = x + 4, {−3, 2}