Find the vertex, axis of symmetry and y-intercept in the equation y-3= -1/2 (x+2)^2

Review your parabola equations... We'll use the standard form of the parabola. Rearranging we get: \[y=\frac{-1}{2}\big(x+2\big)+3 \Rightarrow (h,k)=(2,3) \leftarrow \space the\space vertex\]

Let 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: \[y=a(x-h)^2+k\] 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

\[y=\frac{-1}{2}\big[x+(-2)\big]^2+3\] <-- boy, left off the exponent _again_.

So, 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. \[\sqrt{4}=\pm2\]

Thank you very much! (:

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