1. **State the problem:** Find the acute angle between the curves $y_1 = x^2$ and $y_2 = 2x + 3$ at their point of intersection where $x = 1$. 2. **Find the point of intersection:** Calculate $y_1$ and $y_2$ at $x=1$: $$y_1(1) = 1^2 = 1$$ $$y_2(1) = 2(1) + 3 = 5$$ The curves intersect at $x=1$, but $y_1(1) \neq y_2(1)$, so no intersection at $x=1$. However, the problem asks for the angle at $x=1$ which means the angle between the tangents to the curves at $x=1$. 3. **Find the slopes of the tangents at $x=1$:** For $y_1 = x^2$, derivative is: $$y_1' = 2x$$ At $x=1$: $$m_1 = 2(1) = 2$$ For $y_2 = 2x + 3$, derivative is: $$y_2' = 2$$ At $x=1$: $$m_2 = 2$$ 4. **Formula for angle between two lines with slopes $m_1$ and $m_2$:** $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ 5. **Calculate the angle:** $$\tan \theta = \left| \frac{2 - 2}{1 + 2 \times 2} \right| = \left| \frac{0}{1 + 4} \right| = 0$$ 6. **Find the angle $\theta$:** $$\theta = \arctan(0) = 0^\circ$$ 7. **Interpretation:** The tangents are parallel at $x=1$, so the acute angle between the curves at that point is $0^\circ$. **Final answer:** The acute angle between the curves at $x=1$ is $0^\circ$.