To find the answers to the given quadratic function \( f(x) = 3x^2 - 12x + 11 \), we need to determine its vertex, y-intercept, axis of symmetry, and one of its zeros. Let’s go through these step by step:
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Vertex: The vertex of a quadratic function in the form \( f(x) = ax^2 + bx + c \) 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: \[ f(2) = 3(2)^2 - 12(2) + 11 = 3(4) - 24 + 11 = 12 - 24 + 11 = -1 \] Thus, the vertex is \( (2, -1) \).
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Axis of Symmetry: The axis of symmetry is the x-coordinate of the vertex: \[ x = 2 \]
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Y-intercept: The y-intercept occurs when \( x = 0 \): \[ f(0) = 3(0)^2 - 12(0) + 11 = 11 \] So, the y-intercept is \( (0, 11) \).
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Zeros of the Function: To find the zeros (or x-intercepts), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 3 \), \( b = -12 \), and \( c = 11 \): First, calculate the discriminant: \[ b^2 - 4ac = (-12)^2 - 4(3)(11) = 144 - 132 = 12 \] Now we can find the zeros: \[ x = \frac{12 \pm \sqrt{12}}{6} = \frac{12 \pm 2\sqrt{3}}{6} = 2 \pm \frac{\sqrt{3}}{3} \] Evaluating \( 2 - \frac{\sqrt{3}}{3} \) to get one zero rounded to the nearest tenth: \[ 2 - \frac{\sqrt{3}}{3} \approx 2 - 0.577 = 1.423 \text{ (approximately 1.4 rounded to the nearest tenth)} \]
Now we can fill in the responses:
- Vertex: (2, -1)
- Axis of symmetry: \( x = 2 \)
- Y-intercept: (0, 11)
- One of the zeros: (1.4, 0)
Completing the statements:
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 (0, 11) and one of the zeros is (1.4, 0) rounded to the nearest tenth.