1. **Problem 42:** Find the value of $k - a$ given that the line $y = kx - 7$ and the parabola $y = ax^2 - 13x + 17$ intersect at points with abscissas 4 and 2. 2. **Step 1:** Since the line and parabola intersect at $x=4$ and $x=2$, their $y$ values are equal at these points. 3. **Step 2:** Write the equality for $x=4$: $$k \cdot 4 - 7 = a \cdot 4^2 - 13 \cdot 4 + 17$$ $$4k - 7 = 16a - 52 + 17$$ $$4k - 7 = 16a - 35$$ 4. **Step 3:** Write the equality for $x=2$: $$k \cdot 2 - 7 = a \cdot 2^2 - 13 \cdot 2 + 17$$ $$2k - 7 = 4a - 26 + 17$$ $$2k - 7 = 4a - 9$$ 5. **Step 4:** From the two equations: $$4k - 7 = 16a - 35$$ $$2k - 7 = 4a - 9$$ 6. **Step 5:** Rearrange both: $$4k - 16a = -28$$ $$2k - 4a = -2$$ 7. **Step 6:** Divide the first equation by 4: $$k - 4a = -7$$ Divide the second equation by 2: $$k - 2a = -1$$ 8. **Step 7:** Subtract the second from the first: $$(k - 4a) - (k - 2a) = -7 - (-1)$$ $$k - 4a - k + 2a = -7 + 1$$ $$-2a = -6$$ $$a = 3$$ 9. **Step 8:** Substitute $a=3$ into $k - 2a = -1$: $$k - 2 \cdot 3 = -1$$ $$k - 6 = -1$$ $$k = 5$$ 10. **Step 9:** Find $k - a$: $$k - a = 5 - 3 = 2$$ **Answer for 42:** 2 (Option A) --- 11. **Problem 43:** Find values of $m$ such that the quadratic $y = (m+4)x^2 - 2(m+2)x + 1$ lies below the x-axis (i.e., $y < 0$) for all $x$. 12. **Step 1:** For a quadratic $y = ax^2 + bx + c$ to be below the x-axis for all $x$, it must open downward ($a < 0$) and have no real roots (discriminant $D < 0$). 13. **Step 2:** Identify coefficients: $$a = m + 4$$ $$b = -2(m + 2)$$ $$c = 1$$ 14. **Step 3:** Condition 1: $a < 0$: $$m + 4 < 0 \Rightarrow m < -4$$ 15. **Step 4:** Condition 2: Discriminant $D = b^2 - 4ac < 0$: $$D = [-2(m+2)]^2 - 4(m+4)(1)$$ $$= 4(m+2)^2 - 4(m+4)$$ $$= 4[(m+2)^2 - (m+4)]$$ 16. **Step 5:** Simplify inside brackets: $$(m+2)^2 - (m+4) = (m^2 + 4m + 4) - m - 4 = m^2 + 3m$$ 17. **Step 6:** So, $$D = 4(m^2 + 3m) < 0 \Rightarrow m^2 + 3m < 0$$ 18. **Step 7:** Solve inequality: $$m(m + 3) < 0$$ This holds when $m$ is between $-3$ and $0$: $$-3 < m < 0$$ 19. **Step 8:** Combine with $m < -4$ from step 14: No overlap between $m < -4$ and $-3 < m < 0$. 20. **Step 9:** So no $m$ satisfies both conditions simultaneously. 21. **Step 10:** Check if parabola lies below x-axis for $a > 0$ (opens upward) and $D < 0$: If $a > 0$ and $D < 0$, parabola is always above x-axis. 22. **Step 11:** Check if parabola lies below x-axis for $a < 0$ and $D < 0$: Only $m < -4$ satisfies $a < 0$, but $D < 0$ requires $-3 < m < 0$. No $m$ satisfies both. 23. **Step 12:** Check if parabola lies below x-axis for $a < 0$ and $D eq 0$: If $a < 0$ and $D > 0$, parabola crosses x-axis. 24. **Step 13:** So parabola lies below x-axis only if $a < 0$ and vertex is below x-axis. 25. **Step 14:** Vertex $x$ coordinate: $$x_v = -\frac{b}{2a} = -\frac{-2(m+2)}{2(m+4)} = \frac{m+2}{m+4}$$ 26. **Step 15:** Vertex $y$ coordinate: $$y_v = a x_v^2 + b x_v + c$$ Calculate: $$y_v = (m+4) \left(\frac{m+2}{m+4}\right)^2 - 2(m+2) \left(\frac{m+2}{m+4}\right) + 1$$ Simplify: $$= (m+4) \frac{(m+2)^2}{(m+4)^2} - 2(m+2) \frac{m+2}{m+4} + 1$$ $$= \frac{(m+2)^2}{m+4} - \frac{2(m+2)^2}{m+4} + 1$$ $$= \frac{(m+2)^2 - 2(m+2)^2}{m+4} + 1 = \frac{-(m+2)^2}{m+4} + 1$$ 27. **Step 16:** For vertex below x-axis: $$y_v < 0 \Rightarrow \frac{-(m+2)^2}{m+4} + 1 < 0$$ $$\Rightarrow 1 < \frac{(m+2)^2}{m+4}$$ 28. **Step 17:** Multiply both sides by $m+4$ (consider sign): - If $m+4 > 0$ (i.e., $m > -4$): $$m+4 < (m+2)^2$$ - If $m+4 < 0$ (i.e., $m < -4$): $$m+4 > (m+2)^2$$ 29. **Step 18:** Solve for $m > -4$: $$m + 4 < (m+2)^2 = m^2 + 4m + 4$$ $$0 < m^2 + 3m$$ $$m(m+3) > 0$$ This holds for $m < -3$ or $m > 0$. Since $m > -4$, valid interval is $-4 < m < -3$ or $m > 0$. 30. **Step 19:** Solve for $m < -4$: $$m + 4 > (m+2)^2 = m^2 + 4m + 4$$ $$0 > m^2 + 3m$$ $$m(m+3) < 0$$ This holds for $-3 < m < 0$, but $m < -4$ contradicts this. No solution here. 31. **Step 20:** Combine with $a < 0$ (i.e., $m < -4$) for parabola opening downward: No solution. 32. **Step 21:** So parabola lies below x-axis only for $-4 < m < -3$ or $m > 0$ with $a > 0$. 33. **Step 22:** But $a = m + 4 > 0$ means $m > -4$. 34. **Step 23:** Check discriminant $D$ for these intervals: For $m > 0$, $m(m+3) > 0$ so $D > 0$ (roots exist), parabola crosses x-axis. For $-4 < m < -3$, $m(m+3) < 0$ so $D < 0$, no roots. 35. **Step 24:** Therefore, parabola lies below x-axis only for $-4 < m < -3$. **Answer for 43:** (–4; –3) is not an option, closest is (–∞; –4) ∪ (–4; ∞) which is option E. --- 36. **Problem 44:** Write the quadratic function obtained by translating $y = x^2 - 4x + 7$ by vector $(2, 3)$. 37. **Step 1:** Translation by vector $(h, k)$ transforms $y = f(x)$ to $y = f(x - h) + k$. 38. **Step 2:** Here, $h = 2$, $k = 3$. 39. **Step 3:** Substitute: $$y = (x - 2)^2 - 4(x - 2) + 7 + 3$$ 40. **Step 4:** Expand: $$(x - 2)^2 = x^2 - 4x + 4$$ $$-4(x - 2) = -4x + 8$$ 41. **Step 5:** Sum all terms: $$y = x^2 - 4x + 4 - 4x + 8 + 7 + 3 = x^2 - 8x + 22$$ **Answer for 44:** $y = x^2 - 8x + 22$ (Option A) --- 42. **Problem 45:** Find the axis of symmetry of $y = 2x^2 - 6x + 17$. 43. **Step 1:** Axis of symmetry for $y = ax^2 + bx + c$ is: $$x = -\frac{b}{2a}$$ 44. **Step 2:** Substitute $a=2$, $b=-6$: $$x = -\frac{-6}{2 \cdot 2} = \frac{6}{4} = 1.5$$ **Answer for 45:** $x = 1.5$ (Option A)