Question

Find the vertex of the parabola: y = 8x^2 - 4x. Simplify both coordinates and write them as proper fractions, improper fractions, or integers.

Answer 1

To find the vertex of the parabola given by the equation \( y = 8x^2 - 4x \), we can use the vertex formula. The x-coordinate of the vertex for a parabola in the form \( y = ax^2 + bx + c \) is given by:

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

Here, \( a = 8 \) and \( b = -4 \). Plugging these values into the formula gives:

\[
x = -\frac{-4}{2 \cdot 8} = \frac{4}{16} = \frac{1}{4}
\]

Next, we need to find the y-coordinate of the vertex by substituting \( x = \frac{1}{4} \) back into the original equation:

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

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

\[
\left(\frac{1}{4}\right)^2 = \frac{1}{16}
\]

Now substituting back into the equation:

\[
y = 8 \cdot \frac{1}{16} - 4 \cdot \frac{1}{4}
\]

Calculating each term:

\[
8 \cdot \frac{1}{16} = \frac{8}{16} = \frac{1}{2}
\]

and

\[
- 4 \cdot \frac{1}{4} = -1
\]

So now we combine these:

\[
y = \frac{1}{2} - 1 = \frac{1}{2} - \frac{2}{2} = -\frac{1}{2}
\]

Thus, the vertex of the parabola is:

\[
\left( \frac{1}{4}, -\frac{1}{2} \right)
\]

Therefore, the vertex is \(\boxed{\left( \frac{1}{4}, -\frac{1}{2} \right)}\).