You can put this solution on YOUR website!
i need help with this one also
f(x) = (x+2)^2-3
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
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 = approximately -3.7 and -.27
S we plot those points: "(-3,-2) and (-1,-2)
Then draw the parabola: