How do you draw the graph of y = 2x2 − 3x + 2 for -3 <= x <= 4?

The equation is $y = 2x^2 - 3x + 2$.

• For $x = -3$: $y = 2(-3)^2 - 3(-3) + 2 = 2(9) + 9 + 2 = 18 + 9 + 2 = 29$
• For $x = -2$: $y = 2(-2)^2 - 3(-2) + 2 = 2(4) + 6 + 2 = 8 + 6 + 2 = 16$
• For $x = -1$: $y = 2(-1)^2 - 3(-1) + 2 = 2(1) + 3 + 2 = 2 + 3 + 2 = 7$
• For $x = 0$: $y = 2(0)^2 - 3(0) + 2 = 0 - 0 + 2 = 2$
• For $x = 1$: $y = 2(1)^2 - 3(1) + 2 = 2(1) - 3 + 2 = 2 - 3 + 2 = 1$
• For $x = 2$: $y = 2(2)^2 - 3(2) + 2 = 2(4) - 6 + 2 = 8 - 6 + 2 = 4$
• For $x = 3$: $y = 2(3)^2 - 3(3) + 2 = 2(9) - 9 + 2 = 18 - 9 + 2 = 11$
• For $x = 4$: $y = 2(4)^2 - 3(4) + 2 = 2(16) - 12 + 2 = 32 - 12 + 2 = 22$

The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$. For $y = 2x^2 - 3x + 2$, $a=2$ and $b=-3$.

$$x_v = -\frac{-3}{2(2)} = \frac{3}{4} = 0.75$$

Now, substitute $x_v = 0.75$ into the equation to find $y_v$:

$$y_v = 2(0.75)^2 - 3(0.75) + 2 = 2(0.5625) - 2.25 + 2 = 1.125 - 2.25 + 2 = 0.875$$

The vertex is $(0.75, 0.875)$.

To draw the graph:

1. Draw a coordinate plane with the x-axis ranging from at least -3 to 4 and the y-axis ranging from at least 0 to 29.
2. Plot the points from the table: $(-3, 29)$, $(-2, 16)$, $(-1, 7)$, $(0, 2)$, $(1, 1)$, $(2, 4)$, $(3, 11)$, $(4, 22)$.
3. Plot the vertex $(0.75, 0.875)$.
4. Since the coefficient of $x^2$ (which is 2) is positive, the parabola opens upwards. Draw a smooth curve connecting the plotted points, ensuring it passes through the vertex as its lowest point.

The table of values and the vertex are:

$$\boxed{\begin{array}{|c|c|} \hline x & y \\ \hline -3 & 29 \\ -2 & 16 \\ -1 & 7 \\ 0 & 2 \\ 1 & 1 \\ 2 & 4 \\ 3 & 11 \\ 4 & 22 \\ \hline \end{array} \quad \text{Vertex: } (0.75, 0.875)}$$