I can't draw the picture inline here, but the meeting points are easy to find. Where the curves cross, the two y-values are equal, so set x² = 2x + 1 and rearrange to x² − 2x − 1 = 0. The quadratic formula gives x = 1 ± √2, so the curves meet at two points:
- x = 1 − √2 ≈ −0.414, y ≈ 0.172 (the parabola is still just above the line's height here, and the line dips below the parabola on the left side)
- x = 1 + √2 ≈ 2.414, y ≈ 5.828
You can check both: at x = 1 + √2, x² = 3 + 2√2 and 2x + 1 = 3 + 2√2. The same algebra works for any parabola and line, so it's a good pattern for your homework.
To see the graph, this short script plots both curves and marks the crossings. It needs Python with matplotlib installed:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-2, 4, 400)
plt.plot(x, x**2, label='y = x²')
plt.plot(x, 2*x + 1, label='y = 2x + 1')
for r in [1 - np.sqrt(2), 1 + np.sqrt(2)]:
plt.plot(r, r**2, 'ko')
plt.legend()
plt.grid(True)
plt.show()If you're working through the same parabola with someone else, the earlier thread Can you graph y = x²? covers its shape.
