1. **Problem statement:** (a)(i) Express $y = -x^2 - 7x - 15$ in the form $y = -(x + p)^2 + q$. (a)(ii) Sketch the graph of $y = -x^2 - 7x - 15$. (b)(i) Express $y = x^2 - 3x + 4$ in the form $y = (x - p)^2 + q$. (b)(ii) Sketch the graph of $y = x^2 - 3x + 4$. --- 2. **Formula and rules:** To rewrite a quadratic $y = ax^2 + bx + c$ in vertex form, complete the square: $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. For $a < 0$, the parabola opens downward; for $a > 0$, it opens upward. --- 3. **(a)(i) Express $y = -x^2 - 7x - 15$ in vertex form:** Start with: $$y = -x^2 - 7x - 15$$ Factor out $-1$ from the $x$ terms: $$y = -(x^2 + 7x) - 15$$ Complete the square inside the parentheses: Take half of 7: $\frac{7}{2}$, square it: $\left(\frac{7}{2}\right)^2 = \frac{49}{4}$. Add and subtract $\frac{49}{4}$ inside the parentheses: $$y = -\left(x^2 + 7x + \frac{49}{4} - \frac{49}{4}\right) - 15$$ Group the perfect square trinomial: $$y = -\left(\left(x + \frac{7}{2}\right)^2 - \frac{49}{4}\right) - 15$$ Distribute the $-$ sign: $$y = -\left(x + \frac{7}{2}\right)^2 + \frac{49}{4} - 15$$ Convert 15 to quarters: $15 = \frac{60}{4}$ $$y = -\left(x + \frac{7}{2}\right)^2 + \frac{49}{4} - \frac{60}{4}$$ Simplify: $$y = -\left(x + \frac{7}{2}\right)^2 - \frac{11}{4}$$ So, $$p = \frac{7}{2}, \quad q = -\frac{11}{4}$$ --- 4. **(a)(ii) Sketch the graph of $y = -x^2 - 7x - 15$:** - The vertex is at $\left(-\frac{7}{2}, -\frac{11}{4}\right)$. - The parabola opens downward because of the negative leading coefficient. - The axis of symmetry is $x = -\frac{7}{2}$. - The y-intercept is at $y = -15$ when $x=0$. --- 5. **(b)(i) Express $y = x^2 - 3x + 4$ in vertex form:** Start with: $$y = x^2 - 3x + 4$$ Complete the square: Take half of $-3$: $-\frac{3}{2}$, square it: $\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$. Add and subtract $\frac{9}{4}$: $$y = \left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) + 4$$ Group the perfect square: $$y = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} + 4$$ Convert 4 to quarters: $4 = \frac{16}{4}$ $$y = \left(x - \frac{3}{2}\right)^2 + \frac{7}{4}$$ So, $$p = \frac{3}{2}, \quad q = \frac{7}{4}$$ --- 6. **(b)(ii) Sketch the graph of $y = x^2 - 3x + 4$:** - The vertex is at $\left(\frac{3}{2}, \frac{7}{4}\right)$. - The parabola opens upward because the leading coefficient is positive. - The axis of symmetry is $x = \frac{3}{2}$. - The y-intercept is at $y = 4$ when $x=0$. --- **Final answers:** (a)(i) $y = -\left(x + \frac{7}{2}\right)^2 - \frac{11}{4}$ (b)(i) $y = \left(x - \frac{3}{2}\right)^2 + \frac{7}{4}$