1. The quadratic function can have at most two x-intercepts. **True.** A quadratic function is a polynomial of degree 2, so it can have 0, 1, or 2 real x-intercepts. 2. In the equation $y = ax^2 + bx + c$, the value of $c$ represents the x-intercept. **False.** The value $c$ represents the y-intercept, which is the point where the graph crosses the y-axis (when $x=0$). 3. If the value of $a$ is negative in a quadratic equation, the parabola opens upward. **False.** If $a$ is negative, the parabola opens downward. 4. The vertex form of a quadratic function is written as $y = a(x - h)^2 + k$. **True.** This form clearly shows the vertex at $(h, k)$. 5. The process of completing the square can be used to convert a quadratic from standard form to vertex form. **True.** Completing the square rewrites $ax^2 + bx + c$ into vertex form. 6. The quadratic formula can be used to find the vertex of a parabola. **False.** The quadratic formula finds the roots, not the vertex. The vertex can be found using $x = -\frac{b}{2a}$. 7. If the discriminant is negative, the quadratic equation has no real roots. **True.** The discriminant $\Delta = b^2 - 4ac$ determines the nature of roots; negative means no real roots. 8. If $f(x) = x^2 - 3x + 2$, then $f(2) = 0$. Calculate: $$f(2) = 2^2 - 3(2) + 2 = 4 - 6 + 2 = 0$$ **True.** 9. The axis of symmetry of the quadratic function $y = ax^2 + bx + c$ can be found using the formula $x = -b/2x$. **False.** The correct formula is $x = -\frac{b}{2a}$. 10. A quadratic function can never have a negative range. **False.** The range depends on $a$ and the vertex; if $a > 0$, the range is $[k, \infty)$ where $k$ can be negative. --- **SECTION II: APPLICATION** 6.) Convert $y = (x + 3)(x - 2)$ to vertex form. 1. Expand: $$y = x^2 - 2x + 3x - 6 = x^2 + x - 6$$ 2. Complete the square: $$y = x^2 + x - 6$$ Take half of coefficient of $x$: $\frac{1}{2} = 0.5$, square it: $0.5^2 = 0.25$ Add and subtract 0.25: $$y = (x^2 + x + 0.25) - 0.25 - 6 = (x + 0.5)^2 - 6.25$$ **Vertex form:** $$y = (x + 0.5)^2 - 6.25$$ 2.) Factor $y = 9x^2 - 6x + 1$ to find roots. 1. Recognize perfect square trinomial: $$9x^2 - 6x + 1 = (3x - 1)^2$$ 2. Set equal to zero: $$(3x - 1)^2 = 0$$ 3. Solve for $x$: $$3x - 1 = 0 \Rightarrow x = \frac{1}{3}$$ **Root:** $x = \frac{1}{3}$ (double root)