Question

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

Simplify both coordinates and write them as proper fractions, improper fractions, or integers.

Answer 1

To find the vertex of the parabola defined by the equation \( y = x^2 + 3x + 2 \), we can use the vertex formula for a parabola in the standard form \( y = ax^2 + bx + c \), where the vertex \( x \)-coordinate is given by:

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

In this case, the coefficients are:
- \( a = 1 \)
- \( b = 3 \)
- \( c = 2 \)

Substituting these values into the formula for the \( x \)-coordinate of the vertex gives:

\[
x = -\frac{3}{2 \cdot 1} = -\frac{3}{2}
\]

Next, we need to find the \( y \)-coordinate of the vertex by substituting \( x = -\frac{3}{2} \) back into the original equation:

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

Calculating \( \left(-\frac{3}{2}\right)^2 \):

\[
\left(-\frac{3}{2}\right)^2 = \frac{9}{4}
\]

Now calculate \( 3 \left(-\frac{3}{2}\right) \):

\[
3 \left(-\frac{3}{2}\right) = -\frac{9}{2}
\]

We need a common denominator to combine \( \frac{9}{4} \), \( -\frac{9}{2} \), and \( 2 \). The common denominator for \( 4 \) and \( 2 \) is \( 4 \), so we rewrite \( -\frac{9}{2} \) and \( 2 \):

\[
-\frac{9}{2} = -\frac{18}{4}
\]
\[
2 = \frac{8}{4}
\]

Now, we can add these fractions together:

\[
y = \frac{9}{4} - \frac{18}{4} + \frac{8}{4}
\]

Combine the fractions:

\[
y = \frac{9 - 18 + 8}{4} = \frac{-1}{4}
\]

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

\[
\left(-\frac{3}{2}, -\frac{1}{4}\right)
\]

So, the final answer is:

\[
\boxed{\left(-\frac{3}{2}, -\frac{1}{4}\right)}
\]