To analyze the quadratic function \( y = x^2 - 16x + 28 \), we can determine key points such as the vertex, x-intercepts, and y-intercept.
1. Vertex:
The vertex of a parabola given by the standard form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:
\[ x = -\frac{b}{2a} \]
For the equation \( y = x^2 - 16x + 28 \):
- \( a = 1 \)
- \( b = -16 \)
Calculating the x-coordinate of the vertex:
\[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]
To find the y-coordinate of the vertex, substitute \( x = 8 \) back into the equation:
\[ y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \]
Thus, the vertex is at:
\[ \text{Vertex: } (8, -36) \]
2. X-intercepts:
To find the x-intercepts, set \( y = 0 \):
\[ 0 = x^2 - 16x + 28 \]
We can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(28)}}{2(1)} \] \[ x = \frac{16 \pm \sqrt{256 - 112}}{2} \] \[ x = \frac{16 \pm \sqrt{144}}{2} \] \[ x = \frac{16 \pm 12}{2} \]
Calculating the two potential x-values:
- \( x = \frac{16 + 12}{2} = \frac{28}{2} = 14 \)
- \( x = \frac{16 - 12}{2} = \frac{4}{2} = 2 \)
Thus, the x-intercepts are:
\[ \text{X-intercepts: } (14, 0) \text{ and } (2, 0) \]
3. Y-intercept:
To find the y-intercept, set \( x = 0 \):
\[ y = (0)^2 - 16(0) + 28 = 28 \]
Thus, the y-intercept is at:
\[ \text{Y-intercept: } (0, 28) \]
Summary of Key Points:
- Vertex: \( (8, -36) \)
- X-intercepts: \( (14, 0) \) and \( (2, 0) \)
- Y-intercept: \( (0, 28) \)
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