Find the values of x at the point of intersection of the curve \(y = x^{2} + 2x - 3\) and the lines \(y + x = 1\).
Answer Details
To find the point of intersection of the curve and the line, we need to solve the system of equations formed by equating the two equations:
\begin{align*}
y &= x^2 + 2x - 3 \\
y &= -x + 1
\end{align*}
Setting the right-hand sides equal to each other, we get:
\begin{align*}
x^2 + 2x - 3 &= -x + 1 \\
x^2 + 3x - 4 &= 0 \\
(x + 4)(x - 1) &= 0
\end{align*}
Thus, the values of x at the points of intersection are x = -4 and x = 1. To find the corresponding y-values, we substitute these values of x back into either equation. Using the equation y = x^2 + 2x - 3, we get:
\begin{align*}
y &= (-4)^2 + 2(-4) - 3 \\
&= 7
\end{align*}
and
\begin{align*}
y &= (1)^2 + 2(1) - 3 \\
&= 0
\end{align*}
Therefore, the points of intersection are (-4, 7) and (1, 0). So, the correct answer is (1, -4).