Question

Question
Use the elimination method to solve the linear-quadratic system y=x^2−2x−2, y=2x−6.(1 point)
Responses

(2,−2) and (−2,2)
left parenthesis 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis

(−2,−10)
left parenthesis negative 2 comma negative 10 right parenthesis

(2,−2)
left parenthesis 2 comma negative 2 right parenthesis

(−2,−2) and (−2,2)

Answer 1

To solve the given system of equations using the elimination method, we have:

1. \( y = x^2 - 2x - 2 \) (Equation 1)
2. \( y = 2x - 6 \) (Equation 2)

We can set these two equations equal to one another since they both equal \( y \):

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

Now, we can rearrange the equation to one side:

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

This simplifies to:

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

Factoring this quadratic equation gives:

\[
(x - 2)^2 = 0
\]

Thus:

\[
x - 2 = 0 \implies x = 2
\]

Now, we can substitute \( x = 2 \) back into Equation 2 to find the corresponding \( y \):

\[
y = 2(2) - 6 = 4 - 6 = -2
\]

So, we have one solution:

\[
(2, -2)
\]

Next, we can check if there are any other solutions. Since the quadratic \( (x-2)^2 = 0 \) has a double root at \( x = 2 \), we can analyze the function \( y = x^2 - 2x - 2 \):

To find any additional intersections, we substitute \( x = 2 \) back into Equation 1 to confirm:

\[
y = (2)^2 - 2(2) - 2 = 4 - 4 - 2 = -2
\]

It's confirmed again that the solution point is \( (2, -2) \). Next, since the quadratic opens upwards and we only found one intersection, there cannot be any further solutions for this intersection.

Given this analysis, the solution to the linear-quadratic system is:

\[
\boxed{(2, -2)}
\]

None of the other options provided match additional solutions, as only \( (2, -2) \) exists for this problem.