To analyze the quadratic function \( y = x^2 - 16x + 28 \), we can determine the vertex, x-intercepts, and y-intercept.
1. Vertex:
The vertex form of a quadratic equation is given by the formula \( h = -\frac{b}{2a} \) for the x-coordinate, where \( a \) is the coefficient of \( x^2 \) and \( b \) is the coefficient of \( x \).
For our equation:
- \( a = 1 \)
- \( b = -16 \)
Calculating \( h \): \[ h = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]
Now substitute \( x = 8 \) back into the equation to find the y-coordinate: \[ y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \]
Thus, the vertex is at \( (8, -36) \).
2. X-Intercepts:
To find the x-intercepts, set \( y = 0 \) and solve: \[ 0 = x^2 - 16x + 28 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(28)}}{2(1)} = \frac{16 \pm \sqrt{256 - 112}}{2} \] \[ = \frac{16 \pm \sqrt{144}}{2} = \frac{16 \pm 12}{2} \]
Calculating the two x-values:
- \( x = \frac{28}{2} = 14 \)
- \( x = \frac{4}{2} = 2 \)
Thus, the x-intercepts are \( (2, 0) \) and \( (14, 0) \).
3. Y-Intercept:
To find the y-intercept, set \( x = 0 \): \[ y = (0)^2 - 16(0) + 28 = 28 \]
Thus, the y-intercept is \( (0, 28) \).
Summary of Key Points:
- Vertex: \( (8, -36) \)
- Smaller x-intercept: \( (2, 0) \)
- Larger x-intercept: \( (14, 0) \)
- Y-intercept: \( (0, 28) \)
Final Answers:
- vertex: (8, -36)
- (smaller) x-intercept: (2, 0)
- (larger) x-intercept: (14, 0)
- y-intercept: (0, 28)