To find the key points on the graph of the quadratic function \( y = x^2 - 16x + 28 \), we need to identify the vertex, x-intercepts, and y-intercept.
1. Vertex
The vertex of a quadratic in standard form \( y = ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] For the given function, \( a = 1 \) and \( b = -16 \): \[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \] Now substituting \( x = 8 \) back into the function to find the y-coordinate of the vertex: \[ y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \] So, the vertex is \( (8, -36) \).
2. X-Intercepts
To find the x-intercepts, we set \( y = 0 \): \[ x^2 - 16x + 28 = 0 \] We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -16, c = 28 \): \[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(28)}}{2(1)} = \frac{16 \pm \sqrt{256 - 112}}{2} = \frac{16 \pm \sqrt{144}}{2} = \frac{16 \pm 12}{2} \] Calculating the two values: \[ x = \frac{28}{2} = 14 \quad \text{(larger x-intercept)} \] \[ x = \frac{4}{2} = 2 \quad \text{(smaller x-intercept)} \] So, the x-intercepts are \( (2, 0) \) and \( (14, 0) \).
3. Y-Intercept
To find the y-intercept, we set \( x = 0 \): \[ y = (0)^2 - 16(0) + 28 = 28 \] So, 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 answer:
- Vertex: \( (8, -36) \)
- Smaller x-intercept: \( (2, 0) \)
- Larger x-intercept: \( (14, 0) \)
- Y-intercept: \( (0, 28) \)