Question

What are the two equations for y = -6x^2 + 3x + 8?

Answer 1

The equation \( y = -6x^2 + 3x + 8 \) is a quadratic equation, which can be expressed in different forms. Here are two common ways to represent this equation:

1. **Standard Form**: The equation is already in standard form (ax^2 + bx + c format), where:
- \( a = -6 \)
- \( b = 3 \)
- \( c = 8 \)

So in standard form, it remains:
\[
y = -6x^2 + 3x + 8
\]

2. **Vertex Form**: To convert the equation into vertex form, we complete the square. The vertex form of a quadratic equation is:
\[
y = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex of the parabola.

Starting with the original equation:
\[
y = -6x^2 + 3x + 8
\]

1. Factor out \(-6\) from the quadratic terms:
\[
y = -6\left(x^2 - \frac{1}{2}x\right) + 8
\]

2. Complete the square inside the parentheses. Take \(\frac{-1/2}{2} = -\frac{1}{4}\), square it to get \(\frac{1}{16}\), and then adjust the equation:
\[
y = -6\left(x^2 - \frac{1}{2}x + \frac{1}{16} - \frac{1}{16}\right) + 8
\]
\[
y = -6\left(\left(x - \frac{1}{4}\right)^2 - \frac{1}{16}\right) + 8
\]
\[
y = -6\left(x - \frac{1}{4}\right)^2 + \frac{6}{16} + 8
\]
\[
y = -6\left(x - \frac{1}{4}\right)^2 + \frac{3}{8} + \frac{64}{8}
\]
\[
y = -6\left(x - \frac{1}{4}\right)^2 + \frac{67}{8}
\]

Thus, in vertex form, the equation is:
\[
y = -6\left(x - \frac{1}{4}\right)^2 + \frac{67}{8}
\]

In summary, the two forms of the equation \( y = -6x^2 + 3x + 8 \) are:
- Standard form: \( y = -6x^2 + 3x + 8 \)
- Vertex form: \( y = -6\left(x - \frac{1}{4}\right)^2 + \frac{67}{8} \)