+92 Given the function f(x)=1/3x^2-3x+12
rewrite the equation in Intercept form and use it to find the roots of the function
+6248 \(f(x) = \dfrac 1 3 x^2 - 3x + 12 = \\ \dfrac 1 3(x^2 -9x+36 )= \\ \dfrac 1 3\left(\left(x-\dfrac 9 2\right)^2-\dfrac{81}{4}+36\right) =\\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2- \dfrac{27}{4}+12 = \\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2+\dfrac{21}{4}\)
\(f(x)=0\\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2 = -\dfrac{21}{4}\\ \left(x-\dfrac 9 2\right)^2= -\dfrac{63}{4}\\ x-\dfrac 9 2 = \pm i\sqrt{\dfrac{63}{4}}\\ x = \dfrac 9 2 \pm i\dfrac 3 2 \sqrt{7}\\ x = \dfrac 3 2\left(3\pm i \sqrt{7}\right) \)
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