Question

Find the vertex of the parabola y = x^2 − x + 3.


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 \( y = x^2 - x + 3 \), we can use the vertex formula for a parabola in the form \( y = ax^2 + bx + c \).

The x-coordinate of the vertex is given by the formula:

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

In our equation, \( a = 1 \) and \( b = -1 \). Plugging in these values:

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

Next, we need to find the y-coordinate of the vertex. We can do this by substituting \( x = \frac{1}{2} \) back into the original equation:

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

Calculating each term:

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

Combining the fractions:

\[
y = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4}
\]
\[
y = -\frac{1}{4} + 3 = -\frac{1}{4} + \frac{12}{4} = \frac{11}{4}
\]

Thus, the coordinates of the vertex are:

\[
\left(\frac{1}{2}, \frac{11}{4}\right)
\]

Therefore, the vertex of the parabola \( y = x^2 - x + 3 \) is:

\[
\boxed{\left(\frac{1}{2}, \frac{11}{4}\right)}
\]