To solve this question and provide answers for each part step-by-step:

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Part 1: Find the vertex

The function is given as:
$f(x) = 6x^2 + 12x - 5$

The vertex form of a quadratic equation is:
$f(x) = a(x-h)^2 + k$
where $(h, k)$ is the vertex.

To find the vertex, we use the formula for the x-coordinate of the vertex:
$x = -\frac{b}{2a}$
Here, $a = 6$, $b = 12$, and $c = -5$.

Substitute into the formula:
$x = -\frac{12}{2(6)} = -\frac{12}{12} = -1$

Now substitute $x = -1$ into the original function to find the y-coordinate ($f(-1)$):
$f(-1) = 6(-1)^2 + 12(-1) - 5$ $f(-1) = 6(1) - 12 - 5 = 6 - 12 - 5 = -11$

Thus, the vertex is:
$\boxed{(-1, -11)}$

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Part 2: Find the x-intercepts

To find the x-intercepts, set $f(x) = 0$:
$6x^2 + 12x - 5 = 0$
This is a standard quadratic equation, so we solve using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Substitute $a = 6$, $b = 12$, and $c = -5$:
$x = \frac{-12 \pm \sqrt{(12)^2 - 4(6)(-5)}}{2(6)}$ $x = \frac{-12 \pm \sqrt{144 + 120}}{12}$ $x = \frac{-12 \pm \sqrt{264}}{12}$ Simplify $\sqrt{264}$:
$\sqrt{264} \approx 16.25$ Now compute the two roots:
$x = \frac{-12 + 16.25}{12} \quad \text{and} \quad x = \frac{-12 - 16.25}{12}$

1. First root:
   $x = \frac{-12 + 16.25}{12} = \frac{4.25}{12} \approx 0.35$
2. Second root:
   $x = \frac{-12 - 16.25}{12} = \frac{-28.25}{12} \approx -2.35$

Thus, the x-intercepts are approximately:
$\boxed{(0.35, 0), (-2.35, 0)}$

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Part 3: Find the y-intercept

The y-intercept occurs when $x = 0$. Substitute $x = 0$ into $f(x)$:
$f(0) = 6(0)^2 + 12(0) - 5 = -5$

Thus, the y-intercept is:
$\boxed{(0, -5)}$

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Part 4: Axis of symmetry

The axis of symmetry is the vertical line that passes through the vertex. Since the vertex is $(-1, -11)$, the axis of symmetry is:
$\boxed{x = -1}$

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Part 5: Domain and Range

• Domain: Since this is a quadratic function, the domain is all real numbers:
  $\boxed{(-\infty, \infty)}$

• Range: The parabola opens upward (since $a = 6 > 0$), so the minimum value of the function occurs at the vertex. The y-coordinate of the vertex is $-11$, so the range is:
  $\boxed{[-11, \infty)}$

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Part 6: Correct graph

The parabola has:

• Vertex: $(-1, -11)$
• Y-intercept: $(0, -5)$
• X-intercepts: $(0.35, 0)$ and $(-2.35, 0)$

From the options provided, Graph A is correct because:

• The parabola opens upward.
• The vertex is $(-1, -11)$.
• The y-intercept matches $(0, -5)$.

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Final Answers:

1. Vertex: $(-1, -11)$
2. X-intercepts: $(0.35, 0), (-2.35, 0)$
3. Y-intercept: $(0, -5)$
4. Axis of symmetry: $x = -1$
5. Domain: $(-\infty, \infty)$
6. Range: $[-11, \infty)$

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Relative Questions:

1. What does the direction of the parabola depend on in a quadratic equation?
2. How do you determine the vertex of a quadratic function from standard form?
3. What role does the axis of symmetry play in graphing a parabola?
4. How do you find the y-intercept of a function algebraically?
5. How can you verify the solutions of a quadratic equation using a graphing tool?

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Tip: The vertex is always the turning point of the parabola, and its x-coordinate can be found using $x = -\frac{b}{2a}$ for any quadratic in standard form.