Problem 53 - Algebra 2

Question

Write each equation in vertex form. Then identify the vertex, axis of symmetry, and direction of opening. $$ y=-2 x^{2}+16 x-32 $$

Step-by-Step Solution

Answer

Vertex form: \(y = -2(x-4)^2\); Vertex: \((4, 0)\); Axis: \(x=4\); Opens downward.

Step 1: Recognize the Standard Form

The given quadratic equation is \(y = -2x^2 + 16x - 32\). This is in the standard form \(y = ax^2 + bx + c\), where \(a = -2\), \(b = 16\), and \(c = -32\). Our goal is to rewrite this equation in vertex form \(y = a(x - h)^2 + k\).

Step 2: Complete the Square

To rewrite the equation in vertex form, we need to complete the square. First, factor out \(-2\) from the terms that include \(x\): \(y = -2(x^2 - 8x) - 32\). Next, find the value that completes the square inside the parentheses. Take half of the coefficient of \(x\), which is \(-8\), to get \(-4\), and then square it, resulting in \(16\). Add and subtract this square inside the parentheses: \(y = -2(x^2 - 8x + 16 - 16) - 32\).

Step 3: Simplify the Expression

Now, simplify the expression: \(y = -2((x - 4)^2 - 16) - 32\). Distribute the \(-2\) factor: \(y = -2(x - 4)^2 + 32 - 32\). Simplify further: \(y = -2(x - 4)^2\). This is the vertex form of the equation.

Step 4: Identify Vertex, Axis, and Direction

The vertex form \(y = -2(x-4)^2\) indicates that the vertex \((h, k)\) is \((4, 0)\). The axis of symmetry is \(x = 4\). Since \(a = -2\) is negative, the parabola opens downwards.

Key Concepts

Complete the Square

To convert a quadratic equation from standard form to vertex form, the technique of 'completing the square' is essential. Standard form is usually given as \(y = ax^2 + bx + c\). Our goal is to express it in vertex form, which is \(y = a(x - h)^2 + k\) where \((h, k)\) represents the vertex of the parabola.

Here's how completing the square works:

• Identify the quadratic and linear terms. In the equation \(y = -2x^2 + 16x - 32\), focus on \(-2x^2 + 16x\).
• Factor out the leading coefficient (if it's not 1) from these terms, resulting in \(-2(x^2 - 8x)\).
• To complete the square, look at the coefficient of x, which is -8, halve it to get -4, and then square it to get 16.
• Add and subtract this square inside the parenthesis: \(-2(x^2 - 8x + 16 - 16)\). This maintains the equation's balance.

This results in a perfect square trinomial inside the parentheses, allowing you to rewrite it as \((x-4)^2\) and simplifying the equation further until it reaches vertex form.

Vertex Identification

Once the quadratic equation is written in vertex form \(y = a(x - h)^2 + k\), identifying the vertex is straightforward. The vertex of the parabola is given by the coordinates \((h, k)\).

For our equation \(y = -2(x - 4)^2\), the vertex form clarifies the vertex directly. You can easily notice that \(h = 4\) and \(k = 0\). Therefore, the vertex is at point \((4, 0)\).

This vertex provides a critical point of the parabola, representing either the maximum or minimum of the function depending on the direction it opens.

Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola. It divides the parabola into two mirror images. Knowing the vertex coordinates, determining this line becomes extremely simple.

For a parabola in vertex form \(y = a(x - h)^2 + k\), the axis of symmetry can be expressed as \(x = h\). From our vertex \((4, 0)\), the axis of symmetry is simply \(x = 4\).

This line helps in studying the parabola's behavior, as any point on the parabola will have a symmetric counterpart across this axis.

Direction of Opening

The 'direction of opening' describes whether the parabola opens upwards or downwards. It is primarily determined by the coefficient \(a\) in the vertex form of the quadratic function \(y = a(x - h)^2 + k\).

Here's how this works:

• If \(a > 0\), the parabola opens upwards, forming a U-shape.
• If \(a < 0\), as in the equation \(y = -2(x - 4)^2\), the parabola opens downwards, resulting in an inverted U-shape.

In our equation, since \(a = -2\), the parabola opens downward, indicating that the vertex at \((4, 0)\) represents the maximum point of the parabola.