Question

Find the vertex and axis of symmetry of y−36=2(x2+10x)

Answer 1

To find the vertex and axis of symmetry for the given equation, we start with the equation:

\[ y - 36 = 2(x^2 + 10x) \]

First, we can rewrite this equation in a standard form:

1. Expand the right side:
\[
y - 36 = 2x^2 + 20x
\]

2. Rearranging gives:
\[
y = 2x^2 + 20x + 36
\]

Now, we have the equation in the standard quadratic form \( y = ax^2 + bx + c \) where \( a = 2 \), \( b = 20 \), and \( c = 36 \).

### Finding the Vertex

The vertex \( (h, k) \) of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula:

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

Substituting the values of \( a \) and \( b \):

\[
h = -\frac{20}{2 \cdot 2} = -\frac{20}{4} = -5
\]

Next, we find the \( k \) value by substituting \( h \) back into the equation:

\[
k = 2(-5)^2 + 20(-5) + 36
\]
\[
k = 2(25) - 100 + 36
\]
\[
k = 50 - 100 + 36
\]
\[
k = -14
\]

Thus, the vertex is:

\[
\text{Vertex} = (-5, -14)
\]

### Finding the Axis of Symmetry

The axis of symmetry for a parabola in the form \( y = ax^2 + bx + c \) is given by the line \( x = h \).

Since we found that \( h = -5 \), the equation of the axis of symmetry is:

\[
x = -5
\]

### Summary

- **Vertex**: \((-5, -14)\)
- **Axis of Symmetry**: \(x = -5\)