1. **Problem 1: Analyze the function f from its graph** 2. **Determine the domain of definition of f:** The domain is all x-values for which the curve exists. From the description, the curve starts near $x=-2$ and extends to the right beyond $x=4$. Assuming the curve is continuous and defined for all $x$ in this interval, the domain is approximately $$[-2, +\infty)$$. 3. **Number of solutions for equations:** a. For $f(x) = 5$, count the intersections of the curve with the horizontal line $y=5$. Since the maximum is near $x=0$ and the curve rises to a maximum, if the maximum is less than 5, no solutions; if equal to 5, one solution; if greater, two solutions. Without exact values, assume the maximum is less than 5, so **0 solutions**. b. For $f(x) = 1$, count intersections with $y=1$. The curve crosses $y=1$ twice (once going up before max, once going down after max). So, **2 solutions**. 4. **Table of variation of f:** - Increasing from $x=-2$ to $x=0$ (reaching maximum) - Decreasing from $x=0$ to $x=4$ (reaching minimum) - Increasing from $x=4$ onwards 5. **Compare $f(3)$ and $f(3.5)$:** Since the function is decreasing between $0$ and $4$, and $3.5 > 3$, then $$f(3.5) < f(3)$$. 6. **Solve graphically inequalities:** a. $f(x) \leq 0$: Find intervals where the curve is on or below the x-axis. From description, curve crosses x-axis near $x=-2$, $x$ between $0$ and $4$, and again after $4$. So intervals where $f(x) \leq 0$ are approximately $$[-2, a] \cup [b, c]$$ where $a,b,c$ are zeros near described points. b. $f(x) > 0$: Intervals where curve is above x-axis, complementary to above. 7. **Determine $f(4)$ and $f(0)$:** - $f(0)$ is the maximum value of the function. - $f(4)$ is the minimum value. These values represent the local maximum and minimum of the function respectively. --- 8. **Problem 2: Lines and points** 9. a) Show $A(2,1)$ belongs to $(d1)$: Substitute $t$ such that $x = t + 2 = 2 \Rightarrow t=0$. Then $y = 2t + 1 = 2\times0 + 1 = 1$, matches $A$. Show $B(-1,0)$ belongs to $(d2)$: Substitute $x=-1$, $y=0$ into $x + 2y + 1 = -1 + 0 + 1 = 0$, true. 10. b) Show $(d1)$ and $(d2)$ are perpendicular: - Direction vector of $(d1)$ is $(1,2)$ from parametric form. - Normal vector of $(d2)$ is $(1,2)$ from equation $x + 2y + 1=0$. Since direction vector of $(d1)$ equals normal vector of $(d2)$, lines are perpendicular. Show $C(1,-1)$ is intersection: Substitute $x=1$, $y=-1$ into $(d2)$: $1 + 2(-1) + 1 = 1 - 2 + 1 = 0$, true. For $(d1)$, find $t$ such that $x = t + 2 = 1 \Rightarrow t = -1$. Then $y = 2(-1) + 1 = -2 + 1 = -1$, matches $C$. 11. c) Calculate area of triangle $ABC$: Use coordinates: $$A(2,1), B(-1,0), C(1,-1)$$ Area formula: $$\text{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$ Substitute: $$= \frac{1}{2} |2(0 + 1) + (-1)(-1 - 1) + 1(1 - 0)|$$ $$= \frac{1}{2} |2(1) + (-1)(-2) + 1(1)| = \frac{1}{2} |2 + 2 + 1| = \frac{1}{2} \times 5 = 2.5$$ Distance from $C$ to line $(AB)$: Equation of $(AB)$: Slope: $$m = \frac{y_B - y_A}{x_B - x_A} = \frac{0 - 1}{-1 - 2} = \frac{-1}{-3} = \frac{1}{3}$$ Equation: $$y - 1 = \frac{1}{3}(x - 2) \Rightarrow y = \frac{1}{3}x + \frac{1}{3}$$ Distance formula from point $(x_0,y_0)$ to line $Ax + By + C = 0$: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ Rewrite line: $$y - \frac{1}{3}x - \frac{1}{3} = 0 \Rightarrow -\frac{1}{3}x + y - \frac{1}{3} = 0$$ Multiply by 3: $$-x + 3y - 1 = 0$$ Substitute $C(1,-1)$: $$d = \frac{|-1 + 3(-1) - 1|}{\sqrt{(-1)^2 + 3^2}} = \frac{|-1 - 3 - 1|}{\sqrt{1 + 9}} = \frac{|-5|}{\sqrt{10}} = \frac{5}{\sqrt{10}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2}$$ 12. d) Equation of line $(D)$ parallel to $(d2)$ passing through $I(4,5)$: $(d2)$: $x + 2y + 1 = 0$ Parallel lines have same normal vector $(1,2)$, so equation: $$x + 2y + c = 0$$ Substitute $I(4,5)$: $$4 + 2 \times 5 + c = 0 \Rightarrow 4 + 10 + c = 0 \Rightarrow c = -14$$ So equation of $(D)$: $$x + 2y - 14 = 0$$ 13. e) Distance between $(d2)$ and $(D)$: Use distance formula between parallel lines: $$d = \frac{|c_1 - c_2|}{\sqrt{A^2 + B^2}}$$ For $(d2)$: $x + 2y + 1 = 0$, $c_1 = 1$ For $(D)$: $x + 2y - 14 = 0$, $c_2 = -14$ $$d = \frac{|1 - (-14)|}{\sqrt{1^2 + 2^2}} = \frac{15}{\sqrt{5}} = \frac{15\sqrt{5}}{5} = 3\sqrt{5}$$