To answer the questions about the parabola and the function \( y = x^2 - 16x + 28 \), we can start by determining some key features of the quadratic function.
Question 1
You haven't provided an image, but we can analyze the quadratic function you've given.
For the quadratic equation \( y = x^2 - 16x + 28 \):
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Vertex: The vertex of a parabola given in the standard form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).
In your case:
- \( a = 1 \)
- \( b = -16 \)
So, \[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]
Now to find the y-coordinate of the vertex: \[ y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \]
Thus, the vertex is \( (8, -36) \).
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X-intercepts: Set \( y = 0 \) in the equation: \[ 0 = x^2 - 16x + 28 \] This is a quadratic equation in standard form. We can find the x-intercepts using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -16, c = 28 \).
Compute the discriminant: \[ b^2 - 4ac = (-16)^2 - 4(1)(28) = 256 - 112 = 144 \]
Now find the roots: \[ x = \frac{16 \pm \sqrt{144}}{2} = \frac{16 \pm 12}{2} \] This gives us: \[ x_1 = \frac{28}{2} = 14 \quad \text{(larger x-intercept)} \] \[ x_2 = \frac{4}{2} = 2 \quad \text{(smaller x-intercept)} \]
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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 \( (0, 28) \).
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Axis of symmetry: The axis of symmetry is given by the x-coordinate of the vertex: \[ x = 8 \]
Summary for Question 2:
- Vertex: \( (8, -36) \)
- Smaller x-intercept: \( (2, 0) \)
- Larger x-intercept: \( (14, 0) \)
- y-intercept: \( (0, 28) \)
- Axis of symmetry: \( x = 8 \)
Please let me know if you need any more information or if there are other specific questions!
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