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

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 \(y\)-intercept by evaluating \(f(0)\), which is the constant term in the quadratic expression.
4. Find the \(x\)-intercepts by solving the quadratic equation \(-\frac{3}{2}x^2 - 6x + 6 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
5. State the domain as all real numbers \((-\infty, \infty)\) and determine the range based on the vertex and the parabola opening direction (downward since \(a < 0\)), so the range is \((-\infty, y_{vertex}]\).