Question

What are the vertex and axis of symmetry of the parabola y = x ² − 16x + 63?
*
vertex: (8,−1); axis of symmetry: x = 8
vertex: (8,1); axis of symmetry: x = 8
vertex: (−8,−1); axis of symmetry: x = −8
vertex: (−8,1); axis of symmetry: x = −8

Answer 1

To find the vertex and axis of symmetry of the parabola given by the equation \( y = x^2 - 16x + 63 \), we can use the vertex formula for a parabola in standard form \( y = ax^2 + bx + c \).

1. **Identify the coefficients**:
- \( a = 1 \)
- \( b = -16 \)
- \( c = 63 \)

2. **Find the x-coordinate of the vertex**:
\[
x_{\text{vertex}} = -\frac{b}{2a} = -\frac{-16}{2 \times 1} = \frac{16}{2} = 8
\]

3. **Find the y-coordinate of the vertex by substituting \( x_{\text{vertex}} \) back into the equation**:
\[
y_{\text{vertex}} = (8)^2 - 16(8) + 63
\]
\[
y_{\text{vertex}} = 64 - 128 + 63 = -1
\]

4. **Combine to find the vertex**:
The vertex is \( (8, -1) \).

5. **Find the axis of symmetry**:
The axis of symmetry is the vertical line that passes through the vertex, which is given by \( x = x_{\text{vertex}} \):
\[
x = 8
\]

So, the vertex is \( (8, -1) \) and the axis of symmetry is \( x = 8 \).

The correct option is:
**vertex: (8,−1); axis of symmetry: x = 8**.