To analyze the quadratic function \(y = x^2 - 16x + 28\), we'll find the vertex, x-intercepts, and y-intercept.
1. Vertex
The vertex of a quadratic function in the form \(y = ax^2 + bx + c\) can be found using the formula:
\[ x = -\frac{b}{2a} \]
Here, \(a = 1\) and \(b = -16\):
\[ x = -\frac{-16}{2 \times 1} = \frac{16}{2} = 8 \]
To find the y-coordinate of the vertex, substitute \(x = 8\) into the original equation:
\[ y = (8)^2 - 16(8) + 28 \] \[ y = 64 - 128 + 28 \] \[ y = -36 \]
So, the vertex is at the point \((8, -36)\).
2. X-Intercepts
To find the x-intercepts, set \(y = 0\):
\[ 0 = x^2 - 16x + 28 \]
Now, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 1\), \(b = -16\), and \(c = 28\).
Calculate the discriminant:
\[ b^2 - 4ac = (-16)^2 - 4(1)(28) = 256 - 112 = 144 \]
Now, apply the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{144}}{2(1)} = \frac{16 \pm 12}{2} \]
This gives us two solutions:
\[ x = \frac{28}{2} = 14 \quad \text{and} \quad x = \frac{4}{2} = 2 \]
Thus, 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
- Vertex: \((8, -36)\)
- X-Intercepts: \((14, 0)\) and \((2, 0)\)
- Y-Intercept: \((0, 28)\)
These are the key points on the graph of the quadratic function \(y = x^2 - 16x + 28\).