Question

Given the table of values, complete the table for the quadratic equation y = -6x^2 + 3x + 8 to find the values of the high or low point of the parabola.

Answer 1

To find the high or low point (the vertex) of the parabola defined by the quadratic equation \( y = -6x^2 + 3x + 8 \), we can use the vertex formula for a parabola given by \( y = ax^2 + bx + c \).

The x-coordinate of the vertex can be found using the formula:

\[
x = -\frac{b}{2a}
\]

where \( a = -6 \) and \( b = 3 \).

Substituting the values:

\[
x = -\frac{3}{2(-6)} = -\frac{3}{-12} = \frac{1}{4}
\]

Next, we substitute this x-value back into the quadratic equation to find the y-coordinate of the vertex:

\[
y = -6\left(\frac{1}{4}\right)^2 + 3\left(\frac{1}{4}\right) + 8
\]

Calculating \( \left(\frac{1}{4}\right)^2 = \frac{1}{16} \):

\[
y = -6\left(\frac{1}{16}\right) + \frac{3}{4} + 8
\]
\[
y = -\frac{6}{16} + \frac{3}{4} + 8
\]
\[
y = -\frac{3}{8} + \frac{3}{4} + 8
\]

To combine the fractions, we can rewrite \( \frac{3}{4} \) with a denominator of 8:

\[
y = -\frac{3}{8} + \frac{6}{8} + \frac{64}{8}
\]
\[
y = \left(-3 + 6 + 64\right) / 8
\]
\[
y = \frac{67}{8}
\]

Thus, the vertex of the parabola, which represents the maximum point (since \( a < 0 \)), is at:

\[
\left( \frac{1}{4}, \frac{67}{8} \right)
\]

If you're looking for how to complete a table with specific x-values using the equation \( y = -6x^2 + 3x + 8 \), you can simply substitute those x-values into the equation to get the corresponding y-values.

If you need more assistance or have specific x-values for the table, let me know!