To analyze the quadratic function \( y = x^2 - 16x + 28 \), we can identify the key points including the vertex, x-intercepts, and y-intercept.
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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, \( a = 1 \) and \( b = -16 \): \[ x = -\frac{-16}{2 \times 1} = \frac{16}{2} = 8 \] To find the y-coordinate of the vertex, substitute \( x = 8 \) back into the equation: \[ y = 8^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \] Thus, the vertex is: \[ \text{Vertex: } (8, -36) \]
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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, \( a = 1, b = -16, c = 28 \): \[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \times 1 \times 28}}{2 \times 1} \] \[ = \frac{16 \pm \sqrt{256 - 112}}{2} \] \[ = \frac{16 \pm \sqrt{144}}{2} \] \[ = \frac{16 \pm 12}{2} \] Solving for the two values: \[ x = \frac{28}{2} = 14 \quad \text{and} \quad x = \frac{4}{2} = 2 \] Therefore, the x-intercepts are: \[ \text{Smaller x-intercept: } (2, 0) \] \[ \text{Larger x-intercept: } (14, 0) \]
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Y-Intercept: To find the y-intercept, set \( x = 0 \): \[ y = 0^2 - 16(0) + 28 = 28 \] Thus, the y-intercept is: \[ \text{Y-intercept: } (0, 28) \]
Putting this all together, we have:
- Vertex: \( (8, -36) \)
- Smaller x-intercept: \( (2, 0) \)
- Larger x-intercept: \( (14, 0) \)
- Y-intercept: \( (0, 28) \)