To solve the system of equations given, which are:
1. \(y = x^2 - 3x + 6\)
2. \(y = 2x + 6\)
We need to find the points where these two equations intersect, meaning where they have the same y-values.
To solve this system, we set the two equations equal to each other:
\[x^2 - 3x + 6 = 2x + 6\]
Next, we simplify the equation:
\[x^2 - 3x + 6 = 2x + 6\]
\[x^2 - 3x = 2x\]
\[x^2 - 5x = 0\]
\[x(x - 5) = 0\]
This equation has solutions when:
1. \(x = 0\)
2. \(x - 5 = 0\) which implies \(x = 5\)
Now, we substitute these x-values back into one of the original equations to find the corresponding y-values. Using the equation \(y = x^2 - 3x + 6\), we find:
1. When \(x = 0\): \(y = (0)^2 - 3(0) + 6 = 6\). So, the point is (0, 6).
2. When \(x = 5\): \(y = (5)^2 - 3(5) + 6 = 25 - 15 + 6 = 16\). So, the point is (5, 16).
Therefore, the correct answer is:
C. (0, 6) and (5, 16)
