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

Question

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

Asked on April 25, 2026Computer Science

This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.

Accepted Answer · 2026-04-25T08:35:03.795Z

Here are the solutions to the questions you sent. 3. Quadratic Function Given the function $y = 2x^2 + 4x + 3$ for $-4 \le x \le 0$. a) Complete the following table: Step 1: 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. Step 1: For a quadratic function $y = ax^2 + bx + c$, the x-coordinate of the vertex is given by $x = \frac{-b}{2a}$. Here, $a=2$, $b=4$, $c=3$. $$ x = \frac{-4}{2(2)} = \frac{-4}{4} = -1 $$ Step 2: 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. Step 1: 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. Step 1: 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. Step 1: The x-intercept occurs when $y=0$. Set the function to zero: $$ 2x^2 + 4x + 3 = 0 $$ Step 2: 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: $$ \begin{cases} y \ge 0 \\ x \ge 0 \\ 2x + 3y \ge 6 \end{cases} $$ Step 1: 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 $(3,0)$. Step 2: 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 $p(x)$. Step 1: 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. Step 1: 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, $p(x) = (x+2)(2x^2 + 7x + 3)$. Step 2: 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 $6$. $$ 2x^2 + 7x + 3 = 2x^2 + 6x + x + 3 $$ $$ = 2x(x+3) + 1(x+3) $$ $$ = (2x+1)(x+3) $$ Step 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 $p(x) = 0$. Step 1: Set the completely factorized form of $p(x)$ equal to zero. $$ (x+2)(2x+1)(x+3) = 0 $$ Step 2: 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 $\boxed{x = -2, x = -\frac{1}{2}, \text{ or } x = -3}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.

Accepted Answer · 2026-04-25T08:35:03.795Z

Here are the solutions to the questions you sent. 3. Quadratic Function Given the function $y = 2x^2 + 4x + 3$ for $-4 \le x \le 0$. a) Complete the following table: Step 1: 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. Step 1: For a quadratic function $y = ax^2 + bx + c$, the x-coordinate of the vertex is given by $x = \frac{-b}{2a}$. Here, $a=2$, $b=4$, $c=3$. $$ x = \frac{-4}{2(2)} = \frac{-4}{4} = -1 $$ Step 2: 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. Step 1: 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. Step 1: 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. Step 1: The x-intercept occurs when $y=0$. Set the function to zero: $$ 2x^2 + 4x + 3 = 0 $$ Step 2: 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: $$ \begin{cases} y \ge 0 \\ x \ge 0 \\ 2x + 3y \ge 6 \end{cases} $$ Step 1: 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 $(3,0)$. Step 2: 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 $p(x)$. Step 1: 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. Step 1: 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, $p(x) = (x+2)(2x^2 + 7x + 3)$. Step 2: 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 $6$. $$ 2x^2 + 7x + 3 = 2x^2 + 6x + x + 3 $$ $$ = 2x(x+3) + 1(x+3) $$ $$ = (2x+1)(x+3) $$ Step 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 $p(x) = 0$. Step 1: Set the completely factorized form of $p(x)$ equal to zero. $$ (x+2)(2x+1)(x+3) = 0 $$ Step 2: 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 $\boxed{x = -2, x = -\frac{1}{2}, \text{ or } x = -3}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.

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