If l is the maximum value of -3x^2+4x+1 and m | vedclass

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(A) For $f(x) = -3x^2+4x+1$,the maximum value $l$ occurs at $x = -b/(2a) = -4/(2 	imes -3) = 2/3$. $l = f(2/3) = -3(4/9) + 4(2/3) + 1 = -4/3 + 8/3 +

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$36x^2-12y^2=49$

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(A) For $f(x) = -3x^2+4x+1$,the maximum value $l$ occurs at $x = -b/(2a) = -4/(2 	imes -3) = 2/3$. $l = f(2/3) = -3(4/9) + 4(2/3) + 1 = -4/3 + 8/3 + 1 = 7/3$. For $g(x) = 3x^2+4x+1$,the minimum value $m$ occurs at $x = -4/(2 	imes 3) = -2/3$. $m = g(-2/3) = 3(4/9) + 4(-2/3) + 1 = 4/3 - 8/3 + 1 = -1/3$. The foci are $(l, 0) = (7/3, 0)$ and $(7m, 0) = (7 	imes -1/3, 0) = (-7/3, 0)$. The center of the hyperbola is $(0, 0)$. The distance between foci is $2ae = 7/3 - (-7/3) = 14/3$. Given $e = 2$,we have $2a(2) = 14/3 \implies 4a = 14/3 \implies a = 7/6$. Also,$b^2 = a^2(e^2-1) = (49/36)(4-1) = (49/36)(3) = 49/12$. The equation of the hyperbola is $x^2/a^2 - y^2/b^2 = 1$. $x^2/(49/36) - y^2/(49/12) = 1 \implies 36x^2/49 - 12y^2/49 = 1 \implies 36x^2 - 12y^2 = 49$.

If $l$ is the maximum value of $-3x^2+4x+1$ and $m$ is the minimum value of $3x^2+4x+1$,then the equation of the hyperbola having foci at $(l, 0)$ and $(7m, 0)$ and eccentricity $e=2$ is

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