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Which statements are true about the graph of the function f(x) = 6x – 4 + x2? Select two options.
The vertex form of the function is f(x) = (x – 2)2 + 2.
The vertex of the function is (–3, –13).
The axis of symmetry for the function is x = 3.
The graph increases over the interval (–3, infinity).
The function does not cross the x-axis.
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True are:
- The vertex of the function is (–3, –13).
- The graph increases over the interval (–3, ∞).
Reason: f(x)=x^2+6x−4 = (x+3)^2 −13, so the vertex is (−3, −13) and the axis is x=−3. Since the parabola opens up, it increases for x>−3. The discriminant 36+16=52>0, so it does cross the x-axis (so the "does not cross" statement is false). The given vertex form (x−2)^2+2 is incorrect.
- The vertex of the function is (–3, –13).
- The graph increases over the interval (–3, ∞).
Reason: f(x)=x^2+6x−4 = (x+3)^2 −13, so the vertex is (−3, −13) and the axis is x=−3. Since the parabola opens up, it increases for x>−3. The discriminant 36+16=52>0, so it does cross the x-axis (so the "does not cross" statement is false). The given vertex form (x−2)^2+2 is incorrect.
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