The problem has three parts. (a) Complete the table of values for the quadratic equation $y = 2x^2 + 5x - 2$ for $x$ ranging from $-4$ to $3$. (b) Draw the graph of $y = 2x^2 + 5x - 2$ for $-4 \le x \le 3$. (c) Use the graph to find the roots of the equation, the minimum value of $y$, and the values of $x$ for which $x$ increases as $y$ increases.

1. Problem Description

The problem has three parts.

(a) Complete the table of values for the quadratic equation $y = 2x^2 + 5x - 2$ for $x$ ranging from $-4$ to $3$.

(b) Draw the graph of $y = 2x^2 + 5x - 2$ for $-4 \le x \le 3$.

(c) Use the graph to find the roots of the equation, the minimum value of $y$, and the values of $x$ for which $x$ increases as $y$ increases.

2. Solution Steps

(a) Completing the table of values:

We are given $y = 2x^2 + 5x - 2$. We need to calculate $y$ for $x = -3, -2, -1, 0, 1, 2$.

For $x = -3$:

$y = 2(-3)^2 + 5(-3) - 2 = 2(9) - 15 - 2 = 18 - 15 - 2 = 1$.

For $x = -2$:

$y = 2(-2)^2 + 5(-2) - 2 = 2(4) - 10 - 2 = 8 - 10 - 2 = -4$.

For $x = -1$:

$y = 2(-1)^2 + 5(-1) - 2 = 2(1) - 5 - 2 = 2 - 5 - 2 = -5$.

For $x = 0$:

$y = 2(0)^2 + 5(0) - 2 = 0 + 0 - 2 = -2$.

For $x = 1$:

$y = 2(1)^2 + 5(1) - 2 = 2 + 5 - 2 = 5$.

For $x = 2$:

$y = 2(2)^2 + 5(2) - 2 = 2(4) + 10 - 2 = 8 + 10 - 2 = 16$.

Therefore, the completed table is:

x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3

---|---|---|---|---|---|---|---|---

y | 10 | 1 | -4 | -5 | -2 | 5 | 16 | 31

(b) Drawing the graph:

We would plot the points $(-4, 10), (-3, 1), (-2, -4), (-1, -5), (0, -2), (1, 5), (2, 16), (3, 31)$ on a coordinate plane and draw a smooth curve through them.

(c) Using the graph:

(i) Roots of the equation:

The roots are the values of $x$ where the graph intersects the x-axis (i.e., where $y=0$). From the values calculated above we know there must be one root between -4 and -3, and one between 0 and

1.

(ii) Minimum value of $y$:

The minimum value of $y$ is the y-coordinate of the vertex of the parabola. Based on the calculations in part (a), the minimum is close to $x=-1$.

To find the exact minimum, complete the square:

$y = 2x^2 + 5x - 2 = 2(x^2 + \frac{5}{2}x) - 2 = 2(x^2 + \frac{5}{2}x + (\frac{5}{4})^2) - 2 - 2(\frac{5}{4})^2 = 2(x+\frac{5}{4})^2 - 2 - \frac{25}{8} = 2(x+\frac{5}{4})^2 - \frac{16+25}{8} = 2(x+\frac{5}{4})^2 - \frac{41}{8}$.

So the minimum value of $y$ is $-\frac{41}{8} = -5.125$, occurring at $x = -\frac{5}{4} = -1.25$.

(iii) Values of $x$ for which $x$ increases as $y$ increases:

This occurs for $x$ values greater than the x-coordinate of the vertex. Therefore, $x > -1.25$.

3. Final Answer

(a) The completed table is:

x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3

---|---|---|---|---|---|---|---|---

y | 10 | 1 | -4 | -5 | -2 | 5 | 16 | 31

(b) Graph of $y = 2x^2 + 5x - 2$ for $-4 \le x \le 3$ (Not included because it's a drawing).

(c)

(i) The roots of the equation are approximately $x = -3.13$ and $x = 0.63$.

(ii) The minimum value of $y$ is $-5.125$.

(iii) Values of $x$ for which $x$ increases as $y$ increases are $x > -1.25$.