Given the function y = 2x² + 4x + 3 for −4 × 0.

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3. Quadratic Function
Given the function $y = 2x^2 + 4x + 3$ for $-4 \le x \le 0$ .

a) Complete the following table:

Substitute each $x$ value into the function $y = 2x^2 + 4x + 3$ to find the corresponding $y$ value.
For $x = -4$ :
$y = 2(-4)^2 + 4(-4) + 3 = 2(16) - 16 + 3 = 32 - 16 + 3 = 19$
For $x = -3$ :
$y = 2(-3)^2 + 4(-3) + 3 = 2(9) - 12 + 3 = 18 - 12 + 3 = 9$
For $x = -2$ :
$y = 2(-2)^2 + 4(-2) + 3 = 2(4) - 8 + 3 = 8 - 8 + 3 = 3$
For $x = -1$ :
$y = 2(-1)^2 + 4(-1) + 3 = 2(1) - 4 + 3 = 2 - 4 + 3 = 1$
For $x = 0$ :
$y = 2(0)^2 + 4(0) + 3 = 0 + 0 + 3 = 3$

The completed table is:
$\begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -3 & -2 & -1 & 0 \\ \hline y & 19 & 9 & 3 & 1 & 3 \\ \hline \end{array}$
b) Find the coordinates of its vertex.
For a quadratic function $y = ax^2 + bx + c$ , the x-coordinate of the vertex is given by $x = \frac{-b}{2a}$ .
Here, $a=2$ , $b=4$ , $c=3$ .
$x = \frac{-4}{2(2)} = \frac{-4}{4} = -1$
Substitute $x = -1$ into the function to find the y-coordinate of the vertex.
$y = 2(-1)^2 + 4(-1) + 3 = 2(1) - 4 + 3 = 2 - 4 + 3 = 1$
The coordinates of the vertex are $\boxed{(-1, 1)}$ .

c) Find its y-intercept.
The y-intercept occurs when $x=0$ .
Substitute $x=0$ into the function:
$y = 2(0)^2 + 4(0) + 3 = 3$
The y-intercept is $\boxed{(0, 3)}$ .

d) Find the equation of the axis of symmetry.
The axis of symmetry is a vertical line that passes through the x-coordinate of the vertex.
From part (b), the x-coordinate of the vertex is $-1$ .
The equation of the axis of symmetry is $\boxed{x = -1}$ .

e) Find its x-intercept.
The x-intercept occurs when $y=0$ .
Set the function to zero:
$2x^2 + 4x + 3 = 0$
Calculate the discriminant $\Delta = b^2 - 4ac$ .
$\Delta = (4)^2 - 4(2)(3) = 16 - 24 = -8$
Since the discriminant $\Delta < 0$ , there are no real roots. Therefore, there are $\boxed{\text{no x-intercepts}}$ .

f) Plot the graph.

To plot the graph, you would use the points from the table in part (a) and the vertex from part (b).
Plot the points: $(-4, 19)$ , $(-3, 9)$ , $(-2, 3)$ , $(-1, 1)$ (vertex), $(0, 3)$ (y-intercept).
Connect these points with a smooth curve to form a parabola opening upwards.

4. System of Inequalities
By shading unwanted regions, show the region that satisfies the given system of inequalities:
$\begin{cases} y \ge 0 \\ x \ge 0 \\ 2x + 3y \ge 6 \end{cases}$
Draw the boundary lines for each inequality.
- For $y \ge 0$ , the boundary line is $y=0$ (the x-axis).
- For $x \ge 0$ , the boundary line is $x=0$ (the y-axis).
- For $2x + 3y \ge 6$ , the boundary line is $2x + 3y = 6$ . To draw $2x + 3y = 6$ :
  
  If $x=0$ , $3y=6 \Rightarrow y=2$ . Point: $(0,2)$ .
  
  If $y=0$ , $2x=6 \Rightarrow x=3$ . Point: $(3,0)$ . Draw a solid line connecting $(0,2)$ and $(3,0)$ .
Identify the unwanted regions for each inequality.
- For $y \ge 0$ : The unwanted region is where $y < 0$ (below the x-axis). Shade below the x-axis.
- For $x \ge 0$ : The unwanted region is where $x < 0$ (to the left of the y-axis). Shade to the left of the y-axis.
- For $2x + 3y \ge 6$ : Test a point, e.g., $(0,0)$ . $2(0) + 3(0) = 0$ , which is not $\ge 6$ . So, $(0,0)$ is in the unwanted region. Shade the region containing the origin (below and to the left of the line $2x+3y=6$ ).
The region that satisfies all three inequalities will be the unshaded region. This region is in the first quadrant, above or on the line $2x+3y=6$ .

5. Polynomial Function
Given $p(x) = 2x^3 + 11x^2 + 17x + 6$ .

a) Show that $x+2$ is a factor of $p(x)$ .
According to the Factor Theorem, if $x+2$ is a factor of $p(x)$ , then $p(-2)$ must be equal to 0.
Substitute $x = -2$ into $p(x)$ :
$p(-2) = 2(-2)^3 + 11(-2)^2 + 17(-2) + 6$ $p(-2) = 2(-8) + 11(4) - 34 + 6$ $p(-2) = -16 + 44 - 34 + 6$ $p(-2) = 28 - 34 + 6$ $p(-2) = -6 + 6$ $p(-2) = 0$
Since $p(-2) = 0$ , $\boxed{x+2 \text{ is a factor of } p(x)}$ .

b) Factorize completely.
Since $x+2$ is a factor, we can use polynomial division or synthetic division to divide $p(x)$ by $(x+2)$ .
Using synthetic division with root $-2$ :
-2 | 2 11 17 6 | -4 -14 -6 ----------------- 2 7 3 0
The quotient is $2x^2 + 7x + 3$ .
So, $p(x) = (x+2)(2x^2 + 7x + 3)$ .
Factorize the quadratic expression $2x^2 + 7x + 3$ .
We look for two numbers that multiply to $2 \times 3 = 6$ and add up to 7. These numbers are 1 and 6.
$2x^2 + 7x + 3 = 2x^2 + 6x + x + 3$ $= 2x(x+3) + 1(x+3)$ $= (2x+1)(x+3)$
Combine the factors to get the complete factorization of $p(x)$ .
$p(x) = \boxed{(x+2)(2x+1)(x+3)}$
c) Solve for $x$ the equation $p(x) = 0$ .
Set the completely factorized form of $p(x)$ equal to zero.
$(x+2)(2x+1)(x+3) = 0$
Set each factor equal to zero and solve for $x$ .
$x+2 = 0 \Rightarrow x = -2$ $2x+1 = 0 \Rightarrow 2x = -1 \Rightarrow x = -\frac{1}{2}$ $x+3 = 0 \Rightarrow x = -3$
The solutions for $x$ are $\boxed{x = -2, x = -\frac{1}{2}, \text{ or } x = -3}$ .

Given the function y = 2x² + 4x + 3 for -4 x 0.

Step-by-step solution

Substitute each $x$ value into the function $y = 2x^2 + 4x + 3$ to find the corresponding $y$ value.

For a quadratic function $y = ax^2 + bx + c$ , the x-coordinate of the vertex is given by $x = \frac{-b}{2a}$ .

Substitute $x = -1$ into the function to find the y-coordinate of the vertex.

The y-intercept occurs when $x=0$ .

The axis of symmetry is a vertical line that passes through the x-coordinate of the vertex.

The x-intercept occurs when $y=0$ .

Calculate the discriminant $\Delta = b^2 - 4ac$ .

Draw the boundary lines for each inequality.

Identify the unwanted regions for each inequality.

According to the Factor Theorem, if $x+2$ is a factor of $p(x)$ , then $p(-2)$ must be equal to 0.

Since $x+2$ is a factor, we can use polynomial division or synthetic division to divide $p(x)$ by $(x+2)$ .

Factorize the quadratic expression $2x^2 + 7x + 3$ .

Combine the factors to get the complete factorization of $p(x)$ .

Set the completely factorized form of $p(x)$ equal to zero.

Set each factor equal to zero and solve for $x$ .

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More Computer Science Questions

Key points of y = 2x² + 4x + 3
Point	x	y
Minimum	−1	1
y-intercept	0	3

Given the function y = 2x² + 4x + 3 for -4 x 0.

Step-by-step solution

Substitute each xxx value into the function y=2x2+4x+3y = 2x^2 + 4x + 3y=2x2+4x+3 to find the corresponding yyy value.

For a quadratic function y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c, the x-coordinate of the vertex is given by x=−b2ax = \frac{-b}{2a}x=2a−b​.

Substitute x=−1x = -1x=−1 into the function to find the y-coordinate of the vertex.

The y-intercept occurs when x=0x=0x=0.

The axis of symmetry is a vertical line that passes through the x-coordinate of the vertex.

The x-intercept occurs when y=0y=0y=0.

Calculate the discriminant Δ=b2−4ac\Delta = b^2 - 4acΔ=b2−4ac.