characteristics of a quadratic Look at this equation \( f(x)=-\frac{1}{3}(x+2)^{2}+3 \) What is the vertex \( \square \) StretchV compress/ none \( \square \) is it refected \( \square \) Vertical Shff \( \square \) Horizontal shiff \( \square \) xintercept \( \square \) \( y \)-intercept \( \square \) Doman \( \square \) Range \( \square \) Max/Mn \( \square \) \( \mathrm{f}(1)= \) \( \square \) As \( x \) goes to einfinty \( \square \) As \( x \) goes to infinity \( \square \)
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The equation you've presented is in vertex form, which makes it easy to identify key features of the quadratic function. The vertex of \( f(x) = -\frac{1}{3}(x+2)^{2}+3 \) is at the point \((-2, 3)\). Since the leading coefficient is negative (\(-\frac{1}{3}\)), the graph of the function opens downwards, indicating that it has a maximum point at the vertex. To find the \(x\)-intercept, set \( f(x) = 0 \) and solve for \(x\). The \(y\)-intercept can be found by evaluating \(f(0)\), which gives you another important point. The domain of any quadratic is all real numbers (\(-\infty, \infty\)), while the range, due to the maximum at the vertex, will be \([-\infty, 3]\). As \(x\) approaches infinity, \(f(x)\) decreases without bound because the parabola opens down. Thus, you'll find that the function has a maximum value of 3 with the points you've specified filled out accordingly!