The given equation is:

$y = -3(x + 4)^2 - 7$

Vertex Form:

The equation is already in vertex form, which is:

$y = a(x - h)^2 + k$

In this case:

• $a = -3$
• $h = -4$ (because of the form $x + 4$, which can be written as $x - (-4)$)
• $k = -7$

So, the vertex is at $(-4, -7)$, and the equation is already in vertex form.

Standard Form:

To convert the equation into standard form, we need to expand the squared term and simplify.

$y = -3(x + 4)^2 - 7$

First, expand $(x + 4)^2$:

$(x + 4)^2 = x^2 + 8x + 16$

Now substitute this back into the equation:

$y = -3(x^2 + 8x + 16) - 7$

Distribute the $-3$:

$y = -3x^2 - 24x - 48 - 7$

Combine like terms:

$y = -3x^2 - 24x - 55$

Thus, the equation in standard form is:

$y = -3x^2 - 24x - 55$

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How do you derive the vertex from the vertex form equation?
2. How does changing the value of $a$ affect the graph of a parabola?
3. What is the difference between the vertex form and standard form of a quadratic equation?
4. How would you find the x-intercepts of this quadratic function?
5. Can you convert the standard form back to vertex form?

Tip: The vertex form of a quadratic equation is useful for identifying the vertex of a parabola quickly, which tells you the maximum or minimum point on the graph.