- Identify the x-coordinates from the factored quadratic equation: \(x = -1\) and \(x = -3\).
- Substitute each x-value into the linear equation \(y = 2x + 11\) to find the corresponding y-values:
- For \(x = -1\), \(y = 2(-1) + 11 = 9\).
- For \(x = -3\), \(y = 2(-3) + 11 = 5\).
- Conclude that the intersection points are \((-1, 9)\) and \((-3, 5)\).
- The solution is \(\boxed{(-3, 5) \text{ and } (-1, 9)}\).
Explanation
- Finding the x-coordinates of the intersection
To find the solution set for the system of equations given by \(y = -x^2 - 2x + 8\) and \(y = 2x + 11\), we must identify the coordinates \((x, y)\) where both equations are true. We are already given the first three steps which set the equations equal to each other, rearranged them into a standard quadratic form, and factored the resulting polynomial. The factored equation is \(0 = (x + 1)(x + 3)\). By setting each factor equal to zero, we find the x-coordinates: \(x + 1 = 0 \implies x = -1\) and \(x + 3 = 0 \implies x = -3\).
- Determining the corresponding y-coordinates
Now that we have determined the x-coordinates, \(x = -1\) and \(x = -3\), we need to find the corresponding y-values by substituting these values into one of the original equations. We will use the simpler linear equation, \(y = 2x + 11\), for these calculations.
For \(x = -1\):
\[y = 2(-1) + 11 = -2 + 11 = 9\]
So, one solution is \((-1, 9)\).
For \(x = -3\):
\[y = 2(-3) + 11 = -6 + 11 = 5\]
So, another solution is \((-3, 5)\).
- Finalizing the solution set
By substituting the calculated x-values back into the linear equation, we have identified two ordered pairs that satisfy both equations in the system: \((-1, 9)\) and \((-3, 5)\). Comparing these results to the provided options, the correct set of solutions is \((-3, 5)\) and \((-1, 9)\). Thus, the correct choice is \(\boxed{(-3, 5) \text{ and } (-1, 9)}\).
Examples
In many real-world scenarios, finding intersection points is essential for optimization. For example, if a business models its production costs as a function of output (a quadratic curve) and its revenue as a linear function of output, the intersection points of these two graphs represent the "break-even" points where the business neither gains nor loses money. Calculating these points allows business owners to understand the range of production that leads to profitability.