Question

12.) \\( f(x) = -3x^2 - 18x - 4 \\) (hint: convert to vertex form)
vertex: \\( \underline{\quad} \\) axis of symmetry: \\( \underline{\quad} \\)
domain: \\( \underline{\quad} \\) (inequality)
range: \\( \underline{\quad} \\) (inequality)

Explanation:

Step1: Factor out the coefficient of \(x^2\)

Given \(f(x) = -3x^2 - 18x - 4\), factor out \(-3\) from the first two terms:
\(f(x) = -3(x^2 + 6x) - 4\)

Step2: Complete the square inside the parentheses

To complete the square for \(x^2 + 6x\), take half of \(6\) (which is \(3\)), square it (\(3^2 = 9\)). Add and subtract \(9\) inside the parentheses (but remember to account for the \(-3\) outside):
\(f(x) = -3(x^2 + 6x + 9 - 9) - 4\)
\(f(x) = -3((x + 3)^2 - 9) - 4\)

Step3: Distribute the \(-3\)

\(f(x) = -3(x + 3)^2 + 27 - 4\)
\(f(x) = -3(x + 3)^2 + 23\)

Step4: Find the vertex

The vertex form of a parabola is \(f(x) = a(x - h)^2 + k\), where the vertex is \((h, k)\). Here, \(h = -3\) and \(k = 23\), so the vertex is \((-3, 23)\).

Step5: Find the axis of symmetry

The axis of symmetry for a parabola in vertex form is \(x = h\), so it is \(x = -3\).

Step6: Find the domain

For a quadratic function, the domain is all real numbers, so in inequality form, it is \(x \in (-\infty, \infty)\) or \( -\infty < x < \infty\).

Step7: Find the range

Since the coefficient of \((x + 3)^2\) is \(-3\) (negative), the parabola opens downward. The maximum value is \(k = 23\), so the range is \(y \leq 23\) or \( -\infty < y \leq 23\).

Answer:

• Vertex: \(\boldsymbol{(-3, 23)}\)
• Axis of Symmetry: \(\boldsymbol{x = -3}\)
• Domain: \(\boldsymbol{-\infty < x < \infty}\) (or \(\boldsymbol{x \in (-\infty, \infty)}\))
• Range: \(\boldsymbol{y \leq 23}\) (or \(\boldsymbol{-\infty < y \leq 23}\))