Question

Choose ,begin emphasis,all,end emphasis, of the points where the graph of the line 3 x plus 2 y equals 6 and the graph of the parabola y equals x squared minus 4 x plus 3 intersect.

Answer options with 5 options
A.
the point 0 comma 3

B.
the point 3-halves comma 3-fourths

C.
the point 5-halves comma negative 3-fourths

D.
the point 1 comma 3-halves

E.
the point 4 comma negative 3

Answer 1

To find the points of intersection between the line \(3x + 2y = 6\) and the parabola \(y = x^2 - 4x + 3\), we will solve the equations simultaneously.

1. **Rearranging the line equation**:
\[
2y = 6 - 3x \implies y = 3 - \frac{3}{2}x
\]

2. **Substituting into the parabola**:
\[
3 - \frac{3}{2}x = x^2 - 4x + 3
\]
Simplifying this:
\[
-\frac{3}{2}x = x^2 - 4x
\]
Rearranging:
\[
x^2 - 4x + \frac{3}{2}x = 0 \implies x^2 - \frac{5}{2}x = 0
\]
Factoring:
\[
x(x - \frac{5}{2}) = 0
\]
Thus, \(x = 0\) or \(x = \frac{5}{2}\).

3. **Finding corresponding \(y\) values**:
- For \(x = 0\):
\[
y = 3 - \frac{3}{2}(0) = 3
\]
Thus, one point is \((0, 3)\).

- For \(x = \frac{5}{2}\):
\[
y = 3 - \frac{3}{2} \cdot \frac{5}{2} = 3 - \frac{15}{4} = 3 - 3.75 = -0.75 = -\frac{3}{4}
\]
Thus, the other point is \(\left(\frac{5}{2}, -\frac{3}{4}\right)\).

4. **Conclusion**:
The points of intersection are \((0, 3)\) and \(\left(\frac{5}{2}, -\frac{3}{4}\right)\).

Thus the correct choices are:
- **A: the point \(0, 3\)**
- **C: the point \(\frac{5}{2}, -\frac{3}{4}\)**