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Writing Equations of Parabolas Date_____ Period____ Use the information provided to write the vertex form equation of each parabola. 1) Vertex at origin, Focus: (
Quadratic Equations. This unit is about the solution of quadratic equations. These take the form ax2 +bx+c = 0. We will look at four methods: solution by factorisation, solution by completing the square, solution using a formula, and solution using graphs.
equation has linear terms -2hx and -2ky-they disappear when the center is (0,O). EXAMPLE 1 Find the circle that has a diameter from (1,7) to (5, 7). Solution The center is halfway at (3,7). So r = 2 and (x -3)2+ (y -7)2= 22. EXAMPLE 2 Find the center and radius of the circle x2 -6x + y2 -14y = -54.
Stan Ulam that graph of any quadratic function can be obtained from the core parabola, f~x! 5 x2, by applying basic transformations. We apply terminology from the core parabola to parabolas in general. The point (0, 0) is called thevertex of the core parabola, and they-axis is the axis of symmetry. The axis of symmetry is a help
Example 1. Consider the equation y2 = 4x + 12. Find the coordinates of the focus and the vertex and the equations of the directrix and the axis of symmetry. Graph the equation of the parabola. First, write the equation in the form (y – k)2 = 4p(x – h). y2 = 4x + 12 y2 = 4(x + 3) (y – 0)2 = 4(1)(x + 3) Factor. 4p = 4, so p = 1.
General Equation of a Parabola. Preliminaries Graph of y = x2 Transformation of Graphs. Shifting graphs Stretching graphs Flipping graphs. Objectives Find the equation of a parabola, given the graph. y = x2. ( 2; 4) (2; 4) ( 1; 1) (1; 1) x. Axis of symmetry = y-axis. Vertex at (0; 0) y = x2. ( 2; 4) ( 1; 1) (2; 4) (1; 1) = x2. y.
Find an equation for the parabolic shape of each cable. Complete the table by finding the heights of the suspension cables over the roadway at distances of meters from the center of the bridge. 0. 20.
