Question

given the parabola \\(y = -3x^2 + 12x\\),

a. factor the equation to find the zeros.
b. state the axis of symmetry
c. state the vertex
d. graph label the parabola

Explanation:

Factor the equation to find the zeros

Find the zeros by setting \(y = 0\) and factoring.

$$ -3x^2 + 12x = 0 $$

Factor out the greatest common factor, \(-3x\).

$$ -3x(x - 4) = 0 $$

Set each factor to zero to find the zeros.

$$ x = 0 \quad \text{or} \quad x = 4 $$

State the axis of symmetry

Find the axis of symmetry using the zeros.
The axis of symmetry lies midway between the zeros.

$$ x = \frac{0 + 4}{2} = 2 $$

State the vertex

Find the vertex coordinates \((h, k)\).
The \(x\)-coordinate is the axis of symmetry, \(h = 2\).
Substitute \(x = 2\) into the original equation to find \(y\).

$$ y = -3(2)^2 + 12(2) = -3(4) + 24 = 12 $$

The vertex is \((2, 12)\).

Graph and label the parabola

Identify key points to plot the parabola.

• Vertex: \((2, 12)\)
• Zeros (\(x\)-intercepts): \((0, 0)\) and \((4, 0)\)
• Additional point at \(x = 1\): \(y = -3(1)^2 + 12(1) = 9\), giving point \((1, 9)\)
• Symmetric point at \(x = 3\): \((3, 9)\)

Answer:

Question a

The factored equation is:

$$y = -3x(x - 4)$$

The zeros are:

$$x = 0 \quad \text{and} \quad x = 4$$

Question b

The axis of symmetry is:

$$x = 2$$

Question c

The vertex is:

$$(2, 12)$$

Question d

The graph of the parabola \(y = -3x^2 + 12x\) has its vertex at \((2, 12)\), \(x\)-intercepts at \((0, 0)\) and \((4, 0)\), and passes through \((1, 9)\) and \((3, 9)\).