To analyze the quadratic function \( f(x) = 3x^2 - 12x + 11 \), we will complete the needed statements step by step.
-
Vertex: The vertex of a quadratic function in the form \( f(x) = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).
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 \( (2, -1) \).
-
Axis of Symmetry: The equation of the axis of symmetry for a quadratic function is \( x = -\frac{b}{2a} \). We have already calculated this value: \[ \text{Axis of Symmetry: } x = 2 \]
-
Y-Intercept: The y-intercept of the function can be found by evaluating \( f(0) \): \[ f(0) = 3(0^2) - 12(0) + 11 = 11 \]
So, the y-intercept is \( 11 \).
-
Zeros of the Function: To find the zeros of the function, we set \( f(x) = 0 \): \[ 3x^2 - 12x + 11 = 0 \] We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plugging in the values \( a = 3 \), \( b = -12 \), and \( c = 11 \): \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(11)}}{2(3)} = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = \frac{12 \pm 2\sqrt{3}}{6} \] Simplifying further: \[ x = 2 \pm \frac{\sqrt{3}}{3} \]
To find one of the zeros rounded to the nearest tenth, we can plug in the value \( \sqrt{3} \approx 1.732 \):
- \( x \approx 2 + \frac{1.732}{3} \approx 2 + 0.577 = 2.577 \) rounded to \( 2.6 \),
- \( x \approx 2 - \frac{1.732}{3} \approx 2 - 0.577 = 1.423 \) rounded to \( 1.4 \).
We can use \( 2.6 \) as one of the zeros (you may also use \( 1.4 \) depending on the context).
Summary of 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 2.6 (or \( 1.4 \)).
Using these values, you can fill in the responses in the correct areas as specified.