1. **Problem 34:** Find $q$ for the parabola $y = x^2 + px + q$ given that it touches the $x$-axis at $x=5$. 2. Since the parabola touches the $x$-axis at $x=5$, it means $x=5$ is a root with multiplicity 2 (the parabola is tangent to the axis). 3. The parabola can be written as $y = (x - 5)^2 = x^2 - 10x + 25$. 4. Comparing with $y = x^2 + px + q$, we get $p = -10$ and $q = 25$. 5. Among the options, none is 25, so check if the problem means the parabola passes through $x=5$ on the $x$-axis (i.e., $y=0$ at $x=5$). 6. Substitute $x=5$ into $y = x^2 + px + q$ and set $y=0$: $$0 = 25 + 5p + q$$ 7. We need to find $q$ such that the parabola touches the $x$-axis at $x=5$, so the discriminant must be zero: $$D = p^2 - 4q = 0 \\ q = \frac{p^2}{4}$$ 8. From step 6, $q = -25 - 5p$. 9. Equate the two expressions for $q$: $$\frac{p^2}{4} = -25 - 5p$$ 10. Multiply both sides by 4: $$p^2 = -100 - 20p$$ 11. Rearrange: $$p^2 + 20p + 100 = 0$$ 12. Solve quadratic for $p$: $$p = \frac{-20 \pm \sqrt{400 - 400}}{2} = \frac{-20}{2} = -10$$ 13. Substitute $p = -10$ into $q = -25 - 5p$: $$q = -25 - 5(-10) = -25 + 50 = 25$$ 14. Since $q=25$ is not in options, check if the problem expects $q$ value from options given. Possibly a typo or misinterpretation. --- **Problem 35:** Find values of $a$ such that the vertex of parabola $y = x^2 - 2(a+1)x + 1$ lies below $y=\frac{4}{3}$ and the vertex of $y = ax^2 - x + d$ lies above $y=\frac{4}{3}$. 1. Vertex of $y = x^2 - 2(a+1)x + 1$ is at: $$x_v = \frac{2(a+1)}{2} = a+1$$ 2. Vertex $y$-coordinate: $$y_v = (a+1)^2 - 2(a+1)(a+1) + 1 = (a+1)^2 - 2(a+1)^2 + 1 = - (a+1)^2 + 1$$ 3. Condition vertex below $\frac{4}{3}$: $$- (a+1)^2 + 1 < \frac{4}{3} \\ - (a+1)^2 < \frac{4}{3} - 1 = \frac{1}{3} \\ (a+1)^2 > -\frac{1}{3}$$ 4. Since square is always non-negative, this is always true. 5. For the second parabola $y = ax^2 - x + d$, vertex at: $$x_v = \frac{1}{2a}$$ 6. Vertex $y$-coordinate: $$y_v = a \left(\frac{1}{2a}\right)^2 - \frac{1}{2a} + d = \frac{1}{4a} - \frac{1}{2a} + d = -\frac{1}{4a} + d$$ 7. Condition vertex above $\frac{4}{3}$: $$-\frac{1}{4a} + d > \frac{4}{3}$$ 8. Without $d$ value, cannot solve further. Possibly missing data. --- **Problem 36:** Find real $a$ such that parabola $y_1 = x^2 - x + a$ lies above $y_2 = x^2 - x - a^2$. 1. Subtract: $$y_1 - y_2 = a + a^2$$ 2. For $y_1$ to be above $y_2$, require: $$a + a^2 > 0 \\ a(a+1) > 0$$ 3. This inequality holds when: $$a < -1 \quad \text{or} \quad a > 0$$ --- **Problem 37:** Find $a$ such that graphs of $y = 2ax + 1$ and $y = (a-6)x^2 - 2$ do not intersect. 1. Set equal: $$(a-6)x^2 - 2 = 2ax + 1$$ 2. Rearrange: $$(a-6)x^2 - 2ax - 3 = 0$$ 3. Treat as quadratic in $x$: $$A = a-6, B = -2a, C = -3$$ 4. Discriminant: $$D = B^2 - 4AC = 4a^2 - 4(a-6)(-3) = 4a^2 + 12(a-6) = 4a^2 + 12a - 72$$ 5. For no intersection, require $D < 0$: $$4a^2 + 12a - 72 < 0 \\ a^2 + 3a - 18 < 0$$ 6. Solve quadratic inequality: $$a = \frac{-3 \pm \sqrt{9 + 72}}{2} = \frac{-3 \pm 9}{2}$$ 7. Roots: $$a_1 = 3, a_2 = -6$$ 8. Inequality holds between roots: $$-6 < a < 3$$ --- **Problem 38:** Find $t$ such that $f(x) = 3x^2 + 2tx - (t-1)^2$ satisfies $f(-1) = -2$. 1. Substitute $x = -1$: $$f(-1) = 3(1) + 2t(-1) - (t-1)^2 = 3 - 2t - (t^2 - 2t + 1) = 3 - 2t - t^2 + 2t - 1 = 2 - t^2$$ 2. Set equal to $-2$: $$2 - t^2 = -2 \\ -t^2 = -4 \\ t^2 = 4 \\ t = \pm 2$$ --- **Problem 39:** Find sum of coordinates of intersection points of $y = 4x^2 + 4x + 1$ and $y = 2x + 1$. 1. Set equal: $$4x^2 + 4x + 1 = 2x + 1 \\ 4x^2 + 4x - 2x = 0 \\ 4x^2 + 2x = 0 \\ 2x(2x + 1) = 0$$ 2. Roots: $$x = 0, x = -\frac{1}{2}$$ 3. Find $y$ for each: - For $x=0$, $y=2(0)+1=1$ - For $x=-\frac{1}{2}$, $y=2(-\frac{1}{2})+1=0$ 4. Sum of coordinates: $$(0 + 1) + \left(-\frac{1}{2} + 0\right) = 1 - \frac{1}{2} = \frac{1}{2}$$ --- **Problem 40:** Find range of $y = \sqrt{x - x^2}$. 1. Inside square root must be non-negative: $$x - x^2 \geq 0 \\ x(1 - x) \geq 0$$ 2. This holds for $0 \leq x \leq 1$. 3. Maximum of $x - x^2$ is at $x=\frac{1}{2}$: $$x - x^2 = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$ 4. So range of $y$ is: $$[0, \sqrt{\frac{1}{4}}] = [0, \frac{1}{2}]$$ --- **Problem 41:** Find range of $y = \sqrt{9 - x^2}$. 1. Inside root: $$9 - x^2 \geq 0 \\ -3 \leq x \leq 3$$ 2. Minimum value of $y$ is 0 at $x=\pm 3$. 3. Maximum value is $\sqrt{9} = 3$ at $x=0$. 4. Range is: $$[0, 3]$$