To analyze the quadratic function \(y = x^2 - 16x + 28\), we can find the vertex, x-intercepts, and y-intercept.
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Vertex: The vertex of a quadratic function \(y = ax^2 + bx + c\) can be found using the formula for the x-coordinate, \(x = -\frac{b}{2a}\).
Here, \(a = 1\) and \(b = -16\):
\[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]
Now, substitute \(x = 8\) back into the equation to find the y-coordinate of the vertex:
\[ y = (8)^2 - 16(8) + 28 \] \[ y = 64 - 128 + 28 \] \[ y = -36 \]
So, the vertex is \((8, -36)\).
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X-Intercepts: To find the x-intercepts, set \(y = 0\):
\[ 0 = x^2 - 16x + 28 \]
We can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = -16\), \(c = 28\):
\[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 28}}{2 \cdot 1} \] \[ x = \frac{16 \pm \sqrt{256 - 112}}{2} \] \[ x = \frac{16 \pm \sqrt{144}}{2} \] \[ x = \frac{16 \pm 12}{2} \]
This gives us two solutions:
\[ x_1 = \frac{28}{2} = 14 \quad (\text{larger x-intercept}) \] \[ x_2 = \frac{4}{2} = 2 \quad (\text{smaller x-intercept}) \]
Thus, the x-intercepts are \((2, 0)\) and \((14, 0)\).
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Y-Intercept: To find the y-intercept, 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: \((0, 28)\)
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