Graph the following quadratics and state its vertex, intercepts and domain & range.
$f\left(x\right)=-\frac{3}{2}{x}^2-6x+6$

• A. Vertex: $(-2,12)$; $x$-intercepts: $(0.8,0),(-4.8,0)$; $y$-intercept: $(0,6)$; Domain: $(-\infty ,\infty )$; Range: $(-\infty ,12]$
• B. Vertex: $(2,12)$; $x$-intercepts: $(-0.8,0),(4.8,0)$; $y$-intercept: $(0,6)$; Domain: $(-\infty ,\infty )$; Range: $(-\infty ,12]$
• C. Vertex: $(2,12)$; $x$-intercepts: $(-0.8,0),(4.8,0)$; $y$-intercept: $(0,6)$; Domain: $(-\infty ,\infty )$; Range: $(-\infty ,12]$
• D. Vertex: $(-2,12)$; $x$-intercepts: $(0.8,0),(-4.8,0)$; $y$-intercept: $(0,6)$; Domain: $(-\infty ,\infty )$; Range: $(-\infty ,12]$

Answer: (A)

Verified step by step guidance

1. Identify the quadratic function given: \(f(x) = -\frac{3}{2}x^2 - 6x + 6\).
2. Find the vertex using the vertex formula \(x = -\frac{b}{2a}\), where \(a = -\frac{3}{2}\) and \(b = -6\). Substitute \(x\) back into the function to find the \(y\)-coordinate of the vertex.
3. Calculate the \(x\)-intercepts by setting \(f(x) = 0\) and solving the quadratic equation \(-\frac{3}{2}x^2 - 6x + 6 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
4. Find the \(y\)-intercept by evaluating \(f(0)\), which is the constant term in the quadratic function.
5. State the domain and range: The domain of any quadratic function is all real numbers \((-\infty, \infty)\). Since the parabola opens downward (because \(a < 0\)), the range is \((-\infty, y_{vertex}]\), where \(y_{vertex}\) is the \(y\)-value of the vertex.