To analyze the quadratic equation \( y = x^2 - 16x + 28 \) and determine its key points, we will find the vertex, x-intercepts, and y-intercept.
1. Vertex
The vertex of a quadratic equation in the form \( y = ax^2 + bx + c \) can be found using the formula:
\[ x = -\frac{b}{2a} \]
Here, \( a = 1 \), \( b = -16 \), and \( c = 28 \).
Calculating the x-coordinate of the vertex:
\[ x = -\frac{-16}{2 \times 1} = \frac{16}{2} = 8 \]
Next, we substitute \( x = 8 \) back into the original equation to find the y-coordinate of the vertex:
\[ y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \]
Thus, the vertex is at the point \( (8, -36) \).
2. X-Intercepts
To find the x-intercepts, we set \( y = 0 \) and solve for \( x \):
\[ 0 = x^2 - 16x + 28 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot 28 = 256 - 112 = 144 \]
Now applying the quadratic formula:
\[ x = \frac{16 \pm \sqrt{144}}{2} = \frac{16 \pm 12}{2} \]
Calculating the two solutions:
- \( x = \frac{28}{2} = 14 \)
- \( x = \frac{4}{2} = 2 \)
The x-intercepts are at the points \( (14, 0) \) 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 the point \( (0, 28) \).
Summary of Key Points
- Vertex: \( (8, -36) \)
- X-Intercepts: \( (14, 0) \) and \( (2, 0) \)
- Y-Intercept: \( (0, 28) \)