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anonymous
 5 years ago
Find the vertex, axis of symmetry and yintercept in the equation y3= 1/2 (x+2)^2
anonymous
 5 years ago
Find the vertex, axis of symmetry and yintercept in the equation y3= 1/2 (x+2)^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Review your parabola equations... We'll use the standard form of the parabola. Rearranging we get: \[y=\frac{1}{2}\big(x+2\big)+3 \Rightarrow (h,k)=(2,3) \leftarrow \space the\space vertex\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me show you how I got that (and to slip the exponent in that I forgot): {Aaagh  Vertex form} Here is the VERTEX form of the Parabola: \[y=a(xh)^2+k\] Here's your equation made to match... \[y=\frac{1}{2}\big[x+(2)\big]+3 \] so a=1/2 h=2 and k=3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=\frac{1}{2}\big[x+(2)\big]^2+3\] < boy, left off the exponent _again_.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, again, the vertex is (2,3) (sign error on my first response). The axis of symmetry is h= 2 and the parabola points down {as x goes to infinity,y gets _really_ negative}. To find the y intercepts, substitute 0 for y on the left and solve for x. NOTE: you'll have two value  because the answer to a square root questions is plus or minus. \[\sqrt{4}=\pm2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you very much! (:
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