2.10 Quadratic systems of equations | Algebra II Math | Indiana ... - Free Printable

- The functions are $f(x) = x^2 + 1$ and $g(x) = -x^2 + x + 4$.
- To find the points of intersection, set $f(x) = g(x)$: $x^2 + 1 = -x^2 + x + 4$.
- Rearrange to form a quadratic equation: $2x^2 - x - 3 = 0$.
- Solve using the quadratic formula: $x = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm 5}{4}$.
- The solutions are $x = \frac{3}{2}$ and $x = -1$.
- Substitute back to find y-values: for $x = -1$, $y = (-1)^2 + 1 = 2$; for $x = \frac{3}{2}$, $y = \left(\frac{3}{2}\right)^2 + 1 = \frac{13}{4}$.
- The points of intersection are $(-1, 2)$ and $\left(\frac{3}{2}, \frac{13}{4}\right)$.