# SOLUTION: i need help with this one also f(x) = (x+2)^2-3 also, need to know how to graph the function and find the vertex.

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 Click here to see ALL problems on Quadratic Equations Question 607396: i need help with this one also f(x) = (x+2)^2-3 also, need to know how to graph the function and find the vertex.Answer by Edwin McCravy(8908)   (Show Source): You can put this solution on YOUR website!i need help with this one also f(x) = (x+2)^2-3 ```Learn this. f(x) = a(x-h)² + k has vertex (h,k) and 1 unit right and left of the vertex has a y-coordinate which is "a" units above or below the vertex depending on whether a is positive or negative. So three points on the parabola are (h,k) (h-1,k+a), and (h+1,k+a) Compare your equation f(x) = (x+2)² - 3 or f(x) = 1(x+2)² - 3 to f(x) = a(x-h)² + k And you see that a=1, h=-2 and k=-3 So it has vertex (h,k) = (-2,-3). That was your first equation. Now to graph it: two points on each side of the vertex are (h±1,k+a) = (-2±1,-3+1) = "(-3,-2) and (-1,-2)" We can also get the x-intercepts, that is, the real zeros by setting F(x) = 0 (x+2)² - 3 = 0 (x+2)² = 3 x+2 = x = -2 = x = approximately -3.7 and -.27 S we plot those points: "(-3,-2) and (-1,-2) Then draw the parabola: Edwin```