We are asked to analyze two quadratic equations: $y = x^2 + 8x + 15$ and $y = -2x^2 - 8x - 6$. For each equation, we need to find the x-intercept(s), the y-intercept, and the vertex. Then, we need to sketch the graph using these points.

1. Problem Description

We are asked to analyze two quadratic equations: $y = x^2 + 8x + 15$ and $y = -2x^2 - 8x - 6$. For each equation, we need to find the x-intercept(s), the y-intercept, and the vertex. Then, we need to sketch the graph using these points.

2. Solution Steps

Problem 1: $y = x^2 + 8x + 15$

x-intercept(s): Set $y = 0$ and solve for $x$.

$x^2 + 8x + 15 = 0$

$(x + 3)(x + 5) = 0$

$x = -3$ or $x = -5$

So, the x-intercepts are $(-3, 0)$ and $(-5, 0)$.

y-intercept: Set $x = 0$ and solve for $y$.

$y = (0)^2 + 8(0) + 15$

$y = 15$

So, the y-intercept is $(0, 15)$.

Vertex: The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$, where $a = 1$ and $b = 8$.

$x = -\frac{8}{2(1)} = -4$

Now, find the y-coordinate by plugging $x = -4$ into the equation:

$y = (-4)^2 + 8(-4) + 15 = 16 - 32 + 15 = -1$

So, the vertex is $(-4, -1)$.

Problem 2: $y = -2x^2 - 8x - 6$

x-intercept(s): Set $y = 0$ and solve for $x$.

$-2x^2 - 8x - 6 = 0$

Divide by -2:

$x^2 + 4x + 3 = 0$

$(x + 1)(x + 3) = 0$

$x = -1$ or $x = -3$

So, the x-intercepts are $(-1, 0)$ and $(-3, 0)$.

y-intercept: Set $x = 0$ and solve for $y$.

$y = -2(0)^2 - 8(0) - 6$

$y = -6$

So, the y-intercept is $(0, -6)$.

Vertex: The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$, where $a = -2$ and $b = -8$.

$x = -\frac{-8}{2(-2)} = -\frac{-8}{-4} = -2$

Now, find the y-coordinate by plugging $x = -2$ into the equation:

$y = -2(-2)^2 - 8(-2) - 6 = -2(4) + 16 - 6 = -8 + 16 - 6 = 2$

So, the vertex is $(-2, 2)$.

3. Final Answer

Problem 1:

x-intercepts: $(-3, 0)$ and $(-5, 0)$

y-intercept: $(0, 15)$

Vertex: $(-4, -1)$

Problem 2:

x-intercepts: $(-1, 0)$ and $(-3, 0)$

y-intercept: $(0, -6)$

Vertex: $(-2, 2)$