Question

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an left parenthesis, x, comma, y, right parenthesis(x,y) point.
y, equals, x, squared, plus, 8, x, plus, 11
y=x
2
+8x+11

Answer 1

To find the coordinates of the vertex of the parabola defined by the equation \( y = x^2 + 8x + 11 \), we can use the vertex formula. The x-coordinate of the vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula:

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

In this equation:

- \( a = 1 \) (the coefficient of \( x^2 \))
- \( b = 8 \) (the coefficient of \( x \))
- \( c = 11 \) (the constant term)

Substituting \( a \) and \( b \) into the vertex formula:

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

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by substituting \( x = -4 \) back into the original equation:

\[
y = (-4)^2 + 8(-4) + 11
\]
\[
y = 16 - 32 + 11
\]
\[
y = 16 - 32 + 11 = -5
\]

Thus, the coordinates of the vertex of the parabola are:

\[
\boxed{(-4, -5)}
\]