Use the image to answer the question.

Question

A coordinate plane is shown with x axis from negative 5 to 9 and y axis from negative 5 to 13. A parabola opens downward is a graph of y equals negative x squared minus x plus 12. It intersects with the horizontal line of y equals negative x plus 8.

The graph shows a linear-quadratic system of equations. What is the solution(s) to the system?

(1 point)
Responses

(−2,10) and (2,6)
open paren negative 2 comma 10 close paren and open paren 2 comma 6 close paren

(−4,0) and (3,0) and (8,0)
open paren negative 4 comma 0 close paren and open paren 3 comma 0 close paren and open paren 8 comma 0 close paren

(0,8) and (0,12)
open paren 0 comma 8 close paren and open paren 0 comma 12 close paren

no solution

Answer 1

To solve the linear-quadratic system of equations given by the parabola \(y = -x^2 - x + 12\) and the line \(y = -x + 8\), we set the equations equal to each other to find the points of intersection:

\[ -x^2 - x + 12 = -x + 8 \]

We can simplify this by adding \(x\) and subtracting \(8\) from both sides:

\[ -x^2 + 12 - 8 = 0 \] \[ -x^2 + 4 = 0 \]

Multiplying through by \(-1\):

\[ x^2 - 4 = 0 \]

Factoring gives us:

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

From this, we find:

\[ x = 2 \quad \text{and} \quad x = -2 \]

Now we can substitute these \(x\) values back into either original equation to find the corresponding \(y\) values. Using the line equation \(y = -x + 8\):

1. For \(x = 2\): \[ y = -2 + 8 = 6 \] So one solution is \((2, 6)\).

2. For \(x = -2\): \[ y = -(-2) + 8 = 2 + 8 = 10 \] So the other solution is \((-2, 10)\).

Thus, the solutions to the system are:

\[ (-2, 10) \text{ and } (2, 6) \]

Based on the options provided, the correct response is:

(−2,10) and (2,6)