1. The problem asks which type of equation the quadratic formula can solve. The quadratic formula is used to solve quadratic equations of the form $$ax^2 + bx + c = 0$$. Answer: b.) $$ax^2 + bx + c = 0$$. 2. Find the discriminant of the quadratic $$y = 2x^2 + 3x - 5$$. The discriminant formula is $$\Delta = b^2 - 4ac$$ where $$a=2$$, $$b=3$$, and $$c=-5$$. Calculate: $$3^2 - 4(2)(-5) = 9 + 40 = 49$$. Answer: a.) $$3^2 - 4(2)(-5)$$. 3. If the discriminant is zero, the quadratic has exactly one real root (a repeated root). Answer: b.) One real root (a repeated root). 4. The domain of any quadratic function is all real numbers because you can input any real number into $$x$$. Answer: c.) $$\{x|x\in\mathbb{R}\}$$. 5. For $$y = -2(x - 1)^2 + 5$$, the parabola opens downward (coefficient of squared term is negative), so the maximum value is 5. The range is all $$y$$ such that $$y \leq 5$$. Answer: b.) $$\{y|y \leq 5; y\in\mathbb{R}\}$$. 6. Evaluate $$f(-2)$$ for $$f(x) = x^2 - 4x + 3$$. Calculate: $$(-2)^2 - 4(-2) + 3 = 4 + 8 + 3 = 15$$. Answer: a.) 15. 7. Convert $$y = x^2 + 4x + 1$$ to vertex form. Complete the square: $$y = x^2 + 4x + 1 = (x^2 + 4x + 4) - 4 + 1 = (x + 2)^2 - 3$$. Answer: b.) $$y = (x + 2)^2 - 3$$. 8. Convert $$y = 2x^2 + 5x + 2$$ to factored form. Find factors of $$2 \times 2 = 4$$ that sum to 5: 4 and 1. Rewrite: $$2x^2 + 4x + x + 2 = 2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2)$$. Answer: c.) $$y = (2x + 1)(x + 2)$$.