1. **Problem A: Sketch the graph of the relation $y=1-x^3$.** 2. This is a cubic function where $y$ depends on $x$ as $y=1-x^3$. 3. The general form of a cubic function is $y = ax^3 + bx^2 + cx + d$. Here, $a=-1$, $b=0$, $c=0$, and $d=1$. 4. Important properties: - The graph passes through the point $(0,1)$ because when $x=0$, $y=1$. - As $x$ increases, $x^3$ grows quickly, so $y$ decreases rapidly because of the negative sign. - As $x$ decreases, $x^3$ becomes very negative, so $y$ increases rapidly. 5. To sketch: - Calculate a few points: - $x=-2$, $y=1-(-2)^3=1+8=9$ - $x=-1$, $y=1-(-1)^3=1+1=2$ - $x=0$, $y=1$ - $x=1$, $y=1-1=0$ - $x=2$, $y=1-8=-7$ 6. Plot these points and draw a smooth curve passing through them, showing the cubic shape with a point of inflection near the origin. --- 7. **Problem B: Sketch the graph of the relation defined by inequalities $y < -2x + 2$ and $y > x - 3$.** 8. These inequalities define a region bounded by two lines: - Line 1: $y = -2x + 2$ - Line 2: $y = x - 3$ 9. Important rules: - For $y < -2x + 2$, the region is below the line $y = -2x + 2$. - For $y > x - 3$, the region is above the line $y = x - 3$. 10. To sketch: - Draw the line $y = -2x + 2$ (y-intercept 2, slope -2). - Draw the line $y = x - 3$ (y-intercept -3, slope 1). - Shade the region below the first line and above the second line. 11. The solution is the intersection of these two shaded regions, which forms a wedge-shaped area between the two lines. --- **Final answers:** - For A, the graph is the cubic curve $y=1-x^3$. - For B, the graph is the region between the lines $y=-2x+2$ and $y=x-3$ where $y$ satisfies $y < -2x + 2$ and $y > x - 3$ simultaneously.