Similarly, if we are given an equation of the form y 2 AyBxC=0, we complete the square on the y terms and rewrite in the form (yk) 2 =4p(xh) From this, we should be able to recognize the coordinates of the vertex and the focus as well as the equation of the directrixAlgebra Graph y= (x3)^22 y = (x − 3)2 2 y = ( x 3) 2 2 Find the properties of the given parabola Tap for more steps Use the vertex form, y = a ( x − h) 2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k k a = 1 a = 1 h = 3 h = 3 k = 2 k = 2Given parabola may be written as, (x − 5) 2 (y − 3) 2 = ∣ ∣ ∣ ∣ ∣ 3 2 4 2 3 x − 4 y 1 ∣ ∣ ∣ ∣ ∣ 2 Thus the focus of the parabola is (5, 3) and directrix is, 3 x − 4 y 1 = 0 Therefore the equation of axis of the parabola will perpendicular to the directrix and is given by, 4 x 3 y
Graph A Parabola Whose X Intercepts Are At X 3 And X 5 And Whose Minimum Value Is Y 4 Brainly Com
