Problem 1 - Algebra 2

Question

Write \(y=2 x^{2}-12 x+6\) in standard form. Identify the vertex, axis of symmetry, and direction of opening of the parabola.

Step-by-Step Solution

Answer

Vertex: (3, -12); Axis of Symmetry: x = 3; Opens Upward.

Step 1: Define Standard Form

The standard form of a quadratic equation is \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola. We aim to rewrite our equation \(y = 2x^2 - 12x + 6\) into this form.

Step 2: Complete the Square

To complete the square, we take the equation \(y = 2x^2 - 12x + 6\) and factor out the leading coefficient from the quadratic and linear terms: \(y = 2(x^2 - 6x) + 6\). To complete the square, take half of the \(-6\), square it to get \(9\), and add zero in a clever way: \(y = 2(x^2 - 6x + 9 - 9) + 6\). This simplifies to \(y = 2((x-3)^2 - 9) + 6\).

Step 3: Simplify to Standard Form

Simplify \(y = 2((x-3)^2 - 9) + 6\) to \(y = 2(x-3)^2 - 18 + 6\). Finally, write it as \(y = 2(x-3)^2 - 12\). Now, the equation is in standard form, \(y = a(x-h)^2 + k\), where \(a=2\), \(h=3\), and \(k=-12\).

Step 4: Find the Vertex

From the standard form \(y = 2(x-3)^2 - 12\), we can identify the vertex \((h, k)\) as \((3, -12)\).

Step 5: Determine Axis of Symmetry

The axis of symmetry for a parabola in standard form \(y = a(x-h)^2 + k\) is the vertical line \(x = h\). Thus, the axis of symmetry is \(x = 3\).

Step 6: Identify Direction of Opening

The direction of opening is determined by the coefficient \(a\). If \(a > 0\), the parabola opens upward. Here, \(a = 2 > 0\), so the parabola opens upward.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point and represents either the lowest or highest point on its curve, depending on its orientation. In the standard form of a quadratic equation, which is written as \(y = a(x-h)^2 + k\), the vertex of the parabola is given by the coordinates \((h, k)\). This expression forms when a quadratic equation is rewritten in such a way that showcases its geometric properties.

To identify the vertex in the equation \(y = 2(x-3)^2 - 12\), we observe that the values of \(h\) and \(k\) are \(3\) and \(-12\) respectively. Therefore, the vertex is located at the point \((3, -12)\).

The vertex is not just a point; it represents

• the turning point of the parabola,
• its maximum or minimum value depending on whether it opens upwards or downwards.

Axis of Symmetry of a Parabola

The axis of symmetry of a parabola is a vertical line that slices the parabola into two symmetrical halves. In terms of the quadratic equation in its standard form, \(y = a(x-h)^2 + k\), the axis of symmetry is always given by the equation \(x = h\). This line runs vertically through the vertex and is equidistant from the two mirrored sides of the parabola.

For the equation \(y = 2(x-3)^2 - 12\), we pinpoint the axis of symmetry at \(x = 3\) because \(h = 3\). The axis of symmetry plays a significant role as it helps

• to determine the balance of the parabola,
• in finding points that share the same \(y\)-coordinate by reflecting across this line.

Paying attention to the axis of symmetry can simplify graphing and analyzing properties of quadratic equations.

Completing the Square

Completing the square is a powerful algebraic technique used to transform a quadratic equation into its standard form. This method makes it easier to identify the vertex and axis of symmetry. To complete the square, follow these steps:

1. Start with the quadratic equation \(y = ax^2 + bx + c\). In this example, it's given as \(y = 2x^2 - 12x + 6\).
2. Factor out the leading coefficient from the quadratic and linear terms: \(y = 2(x^2 - 6x) + 6\).
3. Take half of the coefficient of \(x\) (in \(-6\), half is \(-3\)), square it to get \(9\), and then add and subtract this square inside the bracket: \(y = 2(x^2 - 6x + 9 - 9) + 6\).
4. Simplify within the parentheses to make a perfect square trinomial: \(y = 2((x-3)^2 - 9) + 6\).
5. Further simplify to reach the standard form: \(y = 2(x-3)^2 - 12\).

Completing the square not only helps to convert the equation but also reveals important features of the parabola, making analysis and graphing much simpler.