Complete the table 3 for the equation y = x3 + 4x2 + x - 3. Then, use the scale of 2 cm to represent 1 unit on the x-axis and 2cm to represent 5 units on the y-axis. Draw the graph of y = x3 + 4x2 + x - 3. Use your graph to solve the equation x3 + 4x2 + 3/2 x - 4 = 0.

Step 1: Complete Table 3 for the equation $y = x^3 + 4x^2 + x - 3$

For $x = -2$: $y = (-2)^3 + 4(-2)^2 + (-2) - 3$ $y = -8 + 4(4) - 2 - 3$ $y = -8 + 16 - 2 - 3$ $y = 8 - 2 - 3$ $y = 6 - 3$ $y = 3$

For $x = 2$: $y = (2)^3 + 4(2)^2 + (2) - 3$ $y = 8 + 4(4) + 2 - 3$ $y = 8 + 16 + 2 - 3$ $y = 24 + 2 - 3$ $y = 26 - 3$ $y = 23$

Completed Table 3:

Step 2: Draw the graph of $y = x^3 + 4x^2 + x - 3$

• Plot the points from the completed Table 3: $(-4, -7)$, $(-3, 3)$, $(-2, 3)$, $(-1, -1)$, $(0, -3)$, $(1, 3)$, $(2, 23)$.
• Use the given scale: 2 cm represents 1 unit on the x-axis and 2 cm represents 5 units on the y-axis.
• Draw a smooth curve connecting these points to represent the graph of $y = x^3 + 4x^2 + x - 3$.

Step 3: Use the graph to solve the equation $x^3 + 4x^2 + \frac{3}{2}x - 4 = 0$

To solve $x^3 + 4x^2 + \frac{3}{2}x - 4 = 0$ using the graph of $y = x^3 + 4x^2 + x - 3$, we need to rewrite the equation in terms of $y$.

Given equation: $x^3 + 4x^2 + \frac{3}{2}x - 4 = 0$ We know $y = x^3 + 4x^2 + x - 3$. Substitute this into the given equation: $(x^3 + 4x^2 + x - 3) + \frac{3}{2}x - x - 4 + 3 = 0$ $y + \frac{1}{2}x - 1 = 0$ $y = -\frac{1}{2}x + 1$

Now, draw the line $y = -\frac{1}{2}x + 1$ on the same graph. To draw this line, find two points:

• If $x = -4$, $y = -\frac{1}{2}(-4) + 1 = 2 + 1 = 3$. Plot $(-4, 3)$.
• If $x = 0$, $y = -\frac{1}{2}(0) + 1 = 0 + 1 = 1$. Plot $(0, 1)$.
• If $x = 2$, $y = -\frac{1}{2}(2) + 1 = -1 + 1 = 0$. Plot $(2, 0)$.

The solutions to the equation $x^3 + 4x^2 + \frac{3}{2}x - 4 = 0$ are the x-coordinates of the intersection points of the curve $y = x^3 + 4x^2 + x - 3$ and the line $y = -\frac{1}{2}x + 1$.

By observing the graph, the intersection points are approximately:

• $x \approx -3.8$
• $x \approx -1.2$
• $x \approx 1.0$

Solutions: $\mathbf{x \approx -3.8, -1.2, 1.0}$ x \approx -3.8, -1.2, 1.0