1. Problem 1: Solve the quadratic equation $x^2 - 2x = 4$ for $x$. 2. Rewrite the equation in standard form by subtracting 4 from both sides: $$x^2 - 2x - 4 = 0$$ 3. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=-2$, and $c=-4$. 4. Calculate the discriminant: $$\Delta = (-2)^2 - 4(1)(-4) = 4 + 16 = 20$$ 5. Substitute values into the quadratic formula: $$x = \frac{-(-2) \pm \sqrt{20}}{2(1)} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}$$ 6. Final solutions: $$x = 1 + \sqrt{5} \quad \text{or} \quad x = 1 - \sqrt{5}$$ 7. Problem 2: Determine the continuity and identify any discontinuities of the function $f(x) = x^2 \tan x$. 8. The function $f(x) = x^2 \tan x$ is composed of $x^2$, which is continuous everywhere, and $\tan x$, which is discontinuous where $\cos x = 0$. 9. The discontinuities of $\tan x$ occur at $$x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$ 10. Thus, $f(x)$ is discontinuous at these points because $\tan x$ is undefined there. 11. Elsewhere, $f(x)$ is continuous since the product of continuous functions is continuous. 12. Summary: - Equation $x^2 - 2x = 4$ has solutions $x = 1 \pm \sqrt{5}$. - Function $f(x) = x^2 \tan x$ is continuous on $\mathbb{R}$ except at $x = \frac{\pi}{2} + k\pi$, where it has discontinuities. 13. The blue filled square labeled A in the top-right corner of the graph likely represents a point or region related to $f(x)$ near a discontinuity or area of interest.