The problems are to analyze and likely graph the given quadratic equations: 3) $y = -x^2 + 8x - 12$ 4) $y = -x^2 - 8x - 12$ 5) $y = 2x^2 + 4x$ 6) $y = x^2 + 6x + 5$ Due to the limitations, I cannot graph. However, I will provide the vertex and some key points to assist in graphing each equation.

Here are the solutions to the problems, including the problem description, solution steps, and final answer.

1. Problem Description

The problems are to analyze and likely graph the given quadratic equations:

3) $y = -x^2 + 8x - 12$

4) $y = -x^2 - 8x - 12$

5) $y = 2x^2 + 4x$

6) $y = x^2 + 6x + 5$

Due to the limitations, I cannot graph. However, I will provide the vertex and some key points to assist in graphing each equation.

2. Solution Steps

Problem 3: $y = -x^2 + 8x - 12$

Step 1: Find the x-coordinate of the vertex using the formula $x = -b / (2a)$. In this case, $a = -1$ and $b = 8$.

$x = -8 / (2 * -1) = -8 / -2 = 4$

Step 2: Find the y-coordinate of the vertex by plugging the x-coordinate back into the equation.

$y = -(4)^2 + 8(4) - 12 = -16 + 32 - 12 = 4$

Vertex: $(4, 4)$

Step 3: Find the x-intercepts by setting y =

0. $0 = -x^2 + 8x - 12$

$x^2 - 8x + 12 = 0$

$(x - 6)(x - 2) = 0$

$x = 2, 6$

X-intercepts: $(2, 0), (6, 0)$

Step 4: Find the y-intercept by setting x =

0. $y = -(0)^2 + 8(0) - 12 = -12$

Y-intercept: $(0, -12)$

Problem 4: $y = -x^2 - 8x - 12$

Step 1: Find the x-coordinate of the vertex.

$x = -b / (2a) = -(-8) / (2 * -1) = 8 / -2 = -4$

Step 2: Find the y-coordinate of the vertex.

$y = -(-4)^2 - 8(-4) - 12 = -16 + 32 - 12 = 4$

Vertex: $(-4, 4)$

Step 3: Find the x-intercepts.

$0 = -x^2 - 8x - 12$

$x^2 + 8x + 12 = 0$

$(x + 6)(x + 2) = 0$

$x = -6, -2$

X-intercepts: $(-6, 0), (-2, 0)$

Step 4: Find the y-intercept.

$y = -(0)^2 - 8(0) - 12 = -12$

Y-intercept: $(0, -12)$

Problem 5: $y = 2x^2 + 4x$

Step 1: Find the x-coordinate of the vertex.

$x = -b / (2a) = -4 / (2 * 2) = -4 / 4 = -1$

Step 2: Find the y-coordinate of the vertex.

$y = 2(-1)^2 + 4(-1) = 2 - 4 = -2$

Vertex: $(-1, -2)$

Step 3: Find the x-intercepts.

$0 = 2x^2 + 4x$

$0 = 2x(x + 2)$

$x = 0, -2$

X-intercepts: $(0, 0), (-2, 0)$

Step 4: Find the y-intercept.

$y = 2(0)^2 + 4(0) = 0$

Y-intercept: $(0, 0)$

Problem 6: $y = x^2 + 6x + 5$

Step 1: Find the x-coordinate of the vertex.

$x = -b / (2a) = -6 / (2 * 1) = -6 / 2 = -3$

Step 2: Find the y-coordinate of the vertex.

$y = (-3)^2 + 6(-3) + 5 = 9 - 18 + 5 = -4$

Vertex: $(-3, -4)$

Step 3: Find the x-intercepts.

$0 = x^2 + 6x + 5$

$0 = (x + 5)(x + 1)$

$x = -5, -1$

X-intercepts: $(-5, 0), (-1, 0)$

Step 4: Find the y-intercept.

$y = (0)^2 + 6(0) + 5 = 5$

Y-intercept: $(0, 5)$

3. Final Answer

Problem 3: Vertex: $(4, 4)$, X-intercepts: $(2, 0), (6, 0)$, Y-intercept: $(0, -12)$

Problem 4: Vertex: $(-4, 4)$, X-intercepts: $(-6, 0), (-2, 0)$, Y-intercept: $(0, -12)$

Problem 5: Vertex: $(-1, -2)$, X-intercepts: $(0, 0), (-2, 0)$, Y-intercept: $(0, 0)$

Problem 6: Vertex: $(-3, -4)$, X-intercepts: $(-5, 0), (-1, 0)$, Y-intercept: $(0, 5)$