10-x^2=x+4

Crate 2 functand find their intersection points what can be used to solve the equation
g(x)+ x+4, f(x)=10+x^2, {-3,2}
g(x)=x+4, f(x)=10+x^2,{-3,1}
g(x)=x+4, f(x)=10-x^2,{-3,2}

1 answer

To solve the equation \( 10 - x^2 = x + 4 \), we can rearrange it as \( 10 - x^2 - x - 4 = 0 \), leading to:

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

Or by multiplying through by -1:

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

Next, we can factor this quadratic equation:

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

Setting each factor to zero gives us the solutions:

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

Thus, the intersection points of \( f(x) = 10 - x^2 \) and \( g(x) = x + 4 \) are \( x = -3 \) and \( x = 2 \).

Regarding the options you provided, here's the summary:

  • g(x) = x + 4, f(x) = 10 - x^2, {-3, 2}: This matches our problem, as the functions and solution set is correct.

The other options are incorrect because:

  • g(x) = x + 4, f(x) = 10 + x^2, {-3, 1}: Incorrect functions and roots.
  • g(x) = x + 4, f(x) = 10 + x^2, {-3, 2}: Incorrect functions; \( f(x) \) should be \( 10 - x^2 \).

Therefore, the correct pair of functions with their intersection points is:

\[ g(x) = x + 4, f(x) = 10 - x^2, \text{ intersection points } {-3, 2} \]