1. **Problem vii:** Find the value of $h$ such that the vertex of $f(x) = (x + 4)(x - h)$ has a positive $x$-coordinate. 2. **Step 1:** Expand the function: $$f(x) = (x + 4)(x - h) = x^2 - hx + 4x - 4h = x^2 + (4 - h)x - 4h$$ 3. **Step 2:** The vertex $x$-coordinate of a parabola $ax^2 + bx + c$ is given by: $$x = -\frac{b}{2a}$$ Here, $a = 1$, $b = 4 - h$. 4. **Step 3:** Substitute $a$ and $b$: $$x = -\frac{4 - h}{2} = \frac{h - 4}{2}$$ 5. **Step 4:** For the vertex to have a positive $x$-coordinate: $$\frac{h - 4}{2} > 0 \implies h - 4 > 0 \implies h > 4$$ 6. **Step 5:** Check the options: - A: $6 > 4$ (valid) - B: $4 > 4$ (false) - C: $-4 > 4$ (false) - D: $-6 > 4$ (false) **Answer for vii:** $h = 6$ --- 7. **Problem viii:** Find the equation of the parabola after transformation $(x,y) \to (x + 4, -3y + 6)$ applied to $y = x^2$. 8. **Step 1:** Original parabola: $$y = x^2$$ 9. **Step 2:** The transformation means: $$x_{new} = x + 4$$ $$y_{new} = -3y + 6$$ 10. **Step 3:** Express $y$ in terms of $x_{new}$: $$y = (x)^2 = (x_{new} - 4)^2$$ 11. **Step 4:** Substitute into $y_{new}$: $$y_{new} = -3(x_{new} - 4)^2 + 6$$ 12. **Step 5:** Rename $x_{new}$ as $x$ for the function notation: $$f(x) = -3(x - 4)^2 + 6$$ **Answer for viii:** $f(x) = -3(x - 4)^2 + 6$ --- **Summary:** - vii) $h = 6$ - viii) $f(x) = -3(x - 4)^2 + 6$