To analyze the quadratic function \( f(x) = 3x^2 - 12x + 11 \), we need to determine the vertex, the equation of the axis of symmetry, the y-intercept, and the zeros of the function.
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Vertex: The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] Here, \( a = 3 \) and \( b = -12 \): \[ x = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2 \] To find the y-coordinate of the vertex, substitute \( x = 2 \) back into the function: \[ f(2) = 3(2)^2 - 12(2) + 11 = 3(4) - 24 + 11 = 12 - 24 + 11 = -1 \] So the vertex is \( (2, -1) \).
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Axis of Symmetry: The axis of symmetry is given by the x-coordinate of the vertex, which is: \[ x = 2 \]
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Y-Intercept: The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function: \[ f(0) = 3(0)^2 - 12(0) + 11 = 11 \] So the y-intercept is \( (0, 11) \).
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Zeros: To find the zeros of the function (where \( f(x) = 0 \)), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 3 \), \( b = -12 \), \( c = 11 \): \[ b^2 - 4ac = (-12)^2 - 4(3)(11) = 144 - 132 = 12 \] Thus, the zeros are: \[ x = \frac{12 \pm \sqrt{12}}{6} = \frac{12 \pm 2\sqrt{3}}{6} = 2 \pm \frac{\sqrt{3}}{3} \] The approximate values for the zeros (rounded) are about \( 3.3 \) and \( 0.7 \).
To summarize:
- The function \( f(x) = 3x^2 - 12x + 11 \) has a vertex of (2, -1).
- The equation of the axis of symmetry is x = 2.
- The y-intercept of the function is (0, 11).
- One of the zeros of the function is approximately (3.5, 0) or (0.7, 0) depending on which root you are referring to.
Thus, the completed statements are:
The function \( f(x) = 3x^2 - 12x + 11 \) has a vertex of (2,-1), the equation of the axis of symmetry is x=2, the y-intercept of the function is (0, 11), and one of the zeros is approximately (3.5, 0) or (0.7, 0) (check which one you need from the context).
Please confirm the final choice for the zero based on your requirements.
1 answer