To complete the statements about the quadratic function \( f(x) = 3x^2 - 12x + 11 \), we need to find the following:
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Vertex: The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). For this function:
- \( a = 3 \)
- \( b = -12 \)
Calculating the x-coordinate of the vertex: \[ x = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2 \] To find the y-coordinate, substitute \( x = 2 \) into \( f(x) \): \[ f(2) = 3(2^2) - 12(2) + 11 = 3(4) - 24 + 11 = 12 - 24 + 11 = -1 \] Thus, the vertex is \( (2, -1) \).
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Axis of Symmetry: This is the vertical line that passes through the vertex, given by the equation \( x = 2 \).
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Y-Intercept: This is found by evaluating \( f(0) \): \[ f(0) = 3(0^2) - 12(0) + 11 = 11 \] Therefore, the y-intercept is \( 11 \).
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Zeros: To find the zeros of the function, set \( f(x) = 0 \): \[ 3x^2 - 12x + 11 = 0 \] We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] With \( a = 3 \), \( b = -12 \), and \( c = 11 \): \[ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 3 \cdot 11}}{2 \cdot 3} = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = \frac{12 \pm 2\sqrt{3}}{6} = \frac{6 \pm \sqrt{3}}{3} \] The two zeros are: \[ x_1 = 2 + \frac{\sqrt{3}}{3} \quad \text{and} \quad x_2 = 2 - \frac{\sqrt{3}}{3} \] Approximating \( \sqrt{3} \approx 1.732 \): \[ x_1 \approx 2 + 0.577 \approx 2.577, \quad x_2 \approx 2 - 0.577 \approx 1.423 \] Rounding \( x_2 \) to the nearest tenth gives \( 1.4 \).
Final Responses:
- The function \( f(x) = 3x^2 - 12x + 11 \) has a vertex of (2, -1) and x = 2 is the equation of the axis of symmetry.
- The y-intercept of the function is 11 and one of the zeros is 1.4 rounded to the nearest tenth.
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