graph each horizontal parabola and give the domain and range: Place them in the form (y - k)² = 4p(x - h) where the vertex is (h, k), the focus is (h+p, k) the end of the focal chord (or latus rectum) are the points (h+p,k+2p), (h+p, k-2p) the directrix is the vertical line whose equation is x = h-p the axis of symmetry is the horizontal line whose equation is y = k The graph opens to the right when p is positive and to the left when p is negative. The domain is [h, ¥) when it opens right, and (-¥, h] when it opens left. The range is (-¥, ¥) 1. x = y^2 + 2 Get y terms on left, other terms on right: -y² = -x + 2 Multiply through by -1 to make left side positive y² = x - 2 Write y as (y - 0) and x - 2 as 1(x - 2) (y - 0)² = 1(x - 2) Compare to (y - k)² = 4p(x - h) Then h = 2, k = 0, 4p = 1, p = 1/4 = .25 The vertex is (h, k) = (2, 0) The focus is (h+p, k) = (2.25, 0) the end of the focal chord (or latus rectum) are the points (h+p,k+2p), (h+p, k-2p), or (2.25, 0+2·.25) = (2.25, .5) and (2.25, 0-2·.25) = (2.25, -.5) the directrix is the vertical line whose equation is x = h-p, or x = 2-.25 or x = 1.75. the axis of symmetry is the horizontal line whose equation is y = k, or y = 0 (the x-axis) The graph opens to the right because p = .25 is positive The domain is [2, ¥) The range is (-¥, ¥)with the directrix 2. x = -2y^2 + 2y - 3 Get y terms on left, others right: 2y² - 2y = -x -3 Divide through by coefficient of y², which is 2 y² - y = -(1/2)x - 3/2 Add the square of 1/2 the coefficient of y to both sides: The coefficient of y is -1, 1/2 of -1 is -1/2. The square of -1/2 is +1/4 y² - y + 1/4 = -(1/2)x - 3/2 + 1/4 Factor the left side, combine last two terms on the right: (y - 1/2)(y - 1/2) = -(1/2)x - 5/4 Write the left side as the square of a binomial, and factor out -(1/2) on the left: (y - 1/2)² = -1/2(x + 5/2) or we can use decimals if we like: (y - .5)² = -.5(x + 2.5) Compare to (y - k)² = 4p(x - h) Then h = -2.5, k = .5, 4p = -.5, p = -.5/4 = -.125 The vertex is (h, k) = (-2.5, .5) The focus is (h+p, k) = (-2.625, .5) the end of the focal chord (or latus rectum) are the points (h+p,k+2p), (h+p, k-2p), or (-2.625, .5-2·(-.125) ) = (-2.625, .75) and (-2.625, .5-2·(-.125) ) = (2.625, .25) the directrix is the vertical line whose equation is x = h-p, or x = -2.5-(-.125) or x = -2.375. the axis of symmetry is the horizontal line whose equation is y = k, or y = .5 (the x-axis) The graph opens to the left because p = -.125 is negative The domain is (-¥, -2.5] The range is (-¥, ¥) with the directrix and axis of symmetry: 3. x^2 = 1/8y Hey, that's a vertical parabola, not a horizontal one! So it has form (x - h)² = 4p(y - k) where the vertex is (h, k), the focus is (h, k+p) the end of the focal chord (or latus rectum) are the points (h-2p,k+p), (h+2p, k+p) the directrix is the horizontal line whose equation is y = k-p the axis of symmetry is the vertical line whose equation is x = h The graph opens upward when p is positive and downward when p is negative. Its domain is (-¥, ¥) Its range is [k, ¥) when the parabola opens upward and (-¥,k] when it opens downward x^2 = 1/8y Write x as (x-0) and y as (y-0) (x - 0)² = 1/8(y - 0) Compare to (x - h)² = 4p(y - k) Then h = 0, k = 0, 4p = 1/8, p = 1/16 where the vertex is (h, k) or (0, 0), the focus is (0, 0+1/16) or (0, 1/16) the end of the focal chord (or latus rectum) are the points (h-2p,k+p), (h+2p, k+p), or (0-2(1/16), 0+1/16) and (0+2(1/16), 0+1/16), or (-1/8, 1/16) and (1/8, 1/16 the directrix is the horizontal line whose equation is y = k-p or y = -1/16 the axis of symmetry is the vertical line whose equation is x = 0 (the y-axis) The graph opens upward because p = 1/16 which is positive. Its domain is (-¥, ¥) Its range is [0, ¥) with the directrix (it's the green line, and it's so close to the x-axist that you can barely see it: Edwin