We are given the quadratic function:

$f(x) = -4x^2 - 12x + 6$

We will solve for the vertex form, vertex, x-intercepts, and y-intercept step by step.

Step 1: Convert to Vertex Form

The vertex form of a quadratic equation is:

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

where $(h, k)$ is the vertex of the parabola. To convert the given equation into vertex form, we will complete the square.

Completing the Square:

We start with the given quadratic:

$f(x) = -4x^2 - 12x + 6$

Factor out $-4$ from the first two terms:

$f(x) = -4(x^2 + 3x) + 6$

Now, complete the square inside the parentheses. To do this, take half of the coefficient of $x$ (which is $3$), divide it by 2 to get $\frac{3}{2}$, and square it to get $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$.

Add and subtract $\frac{9}{4}$ inside the parentheses:

$f(x) = -4\left(x^2 + 3x + \frac{9}{4} - \frac{9}{4}\right) + 6$

Now simplify the expression:

$f(x) = -4\left(\left(x + \frac{3}{2}\right)^2 - \frac{9}{4}\right) + 6$

Distribute the $-4$:

$f(x) = -4\left(x + \frac{3}{2}\right)^2 + 9 + 6$

Simplify the constants:

$f(x) = -4\left(x + \frac{3}{2}\right)^2 + 15$

So, the vertex form is:

$f(x) = -4\left(x + \frac{3}{2}\right)^2 + 15$

Step 2: Find the Vertex

From the vertex form, we can easily identify the vertex $(h, k)$, which is:

$h = -\frac{3}{2}, \quad k = 15$

Thus, the vertex is:

$\left(-\frac{3}{2}, 15\right)$

Step 3: Find the X-Intercepts

To find the x-intercepts, we set $f(x) = 0$ and solve for $x$.

$0 = -4x^2 - 12x + 6$

This is a quadratic equation, and we will solve it using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -4$, $b = -12$, and $c = 6$.

First, calculate the discriminant:

$\Delta = (-12)^2 - 4(-4)(6) = 144 + 96 = 240$

Now apply the quadratic formula:

$x = \frac{-(-12) \pm \sqrt{240}}{2(-4)} = \frac{12 \pm \sqrt{240}}{-8}$

Simplifying $\sqrt{240}$:

$\sqrt{240} = \sqrt{16 \times 15} = 4\sqrt{15}$

Substitute back into the formula:

$x = \frac{12 \pm 4\sqrt{15}}{-8} = \frac{12}{-8} \pm \frac{4\sqrt{15}}{-8}$

Simplify each term:

$x = -\frac{3}{2} \pm \frac{\sqrt{15}}{2}$

Thus, the two x-intercepts are:

$x_1 = -\frac{3}{2} - \frac{\sqrt{15}}{2}, \quad x_2 = -\frac{3}{2} + \frac{\sqrt{15}}{2}$

These are the exact values of the x-intercepts.

Step 4: Find the Y-Intercept

To find the y-intercept, substitute $x = 0$ into the original equation:

$f(0) = -4(0)^2 - 12(0) + 6 = 6$

Thus, the y-intercept is:

$y = 6$

Summary:

• Vertex form: $f(x) = -4\left(x + \frac{3}{2}\right)^2 + 15$
• Vertex: $\left(-\frac{3}{2}, 15\right)$
• X-intercepts: $x_1 = -\frac{3}{2} - \frac{\sqrt{15}}{2}$, $x_2 = -\frac{3}{2} + \frac{\sqrt{15}}{2}$
• Y-intercept: $y = 6$

Do you need further details or have any questions?

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Here are 5 related questions to deepen your understanding:

1. How does the vertex form of a quadratic function help in identifying its graph?
2. What is the significance of the discriminant in the quadratic formula?
3. How would the graph of this function look based on the vertex and intercepts?
4. How would the function change if the coefficient of $x^2$ were positive?
5. Can you rewrite the function in factored form?

Tip: The vertex is the highest or lowest point of the parabola depending on the sign of $a$. If $a < 0$, the parabola opens downward, making the vertex a maximum point.