) 2 + (y 2. x 1. y c x2 = y

Transcription

1 Graphing Parabola Parabolas A parabola is a set of points P whose distance from a fixed point, called the focus, is equal to the perpendicular distance from P to a line, called the directrix. Since this curve is being defined by distances we need to know the distance formula. d = (x x ) + (y y ).5 By the definition of a parabola, we know FP = PD. Substituting those coordinates into the distance formula, we have 0.5 F(0, c) P(x, y) y = c directrix D(x, c) 0.5 (x 0) + (y c) = (x x) + (y + c) Squaring, x + (y c) = 0 + (y + c) Expanding x + y yc + c = y + yc + c Subtracting c & y x yc = yc x = yc c x = y This is an equation of a parabola with vertex at the origin and c being the distance between the Focus, F, and the origin and the origin and the directrix. Mathematically, we write {(x, y)/! y = } is the graph of a parabola with c x focus F(0, C) and directrix with equation y = c.

2 For our purposes, since we are just graphing equations in the form! y =, c x we will replace! with a different variable; a. so our new equation is: c y = ax Example Graph y = x We can make an xy-chart and plot points. x y 0 0 y = x Example Graph y = x + 3 Again, we can make an xy chart and plot points x y y = x + 3 If we did enough of these graphs, we would notice a few things:

3 First, anytime we have an equation with only one variable squared, we always get this bell shaped curve, a parabola. The very top or bottom of the parabola is called the vertex. So, in Example, the vertex was at (0, 0). In the second example, the the vertex was at (0, 3). The graph is also symmetric with respect the to y-axis (x = 0). And finally, when I add a constant to the quadratic term, that moves the graph up and down. So, to generalize, an equation in the form y = ax + k, the k moves the graph vertically as we can see in the last example, y = x + 3 If the coefficient of the quadratic term, a, is negative, y = x, then we get the same curve as y = x, excepts its reflected across the x axis. Example 3 Graph y = x y = x Let s look at the graph when there is number inside the parentheses being squared. Example Graph y = (x ) By making an xy-chart, we can polo points. x y y = ( x ) 5

4 If you look at this graph, it looks exactly like the graph of y = x, except it has been moved over units to the right. Again, if I do enough of these graphs, I might see a pattern. The graph of an equation y = (x h), moves the graph horizontally. The vertex is at (, 0) Example 5 Graph y = (x + 3) I could make an xy-chart, but Im going to use the pattern of moving the graph 3 units to the left. The vertex is at ( 3, 0) 5 y = ( x + 3) Here s what we need to recognize:. The graph of an equation with only one variable squared is a parabola. The vertex can be found using translations from the parent function, y = x, by moving the graph vertically using k and horizontally using h. 3. The graph is symmetric with respect to x = h y = (x h) + k Equation of a Parabola in Vertex Form V (h, k)

5 Example Graph y = (x + ) + 3 By inspection, the graph is a parabola with vertex (, 3). It opens up because the a > 0. If I pick a convenient value of x, like x = 0, then y =. I can then use symmetry to find a third point to sketch the graph - (, ). f( x) = ( x + ) + 3 x = is the line of symmetry. 5 In Example, we graphed y = (x + ) + 3 and found the vertex was at. (, 3) and line of symmetry at x =. The equation is in vertex form. As can be readily seen, an equation in Vertex Form is written as a perfect square. What would happen of we were given an equation such as y = x +x and asked to graph it? The first thing we should realize is only one of the variables is squared, so the graph of the equation is a parabola. If that s the case, how do we find the vertex? That s right, we have to change the equation into Vertex Form by completing the square. y = x + x y = x + x Given Complete Square y = x + x ( 9) Add & Subtract 9 y = (x + 3) Factor/combine terms The graph is a parabola that opens up with vertex ( 3, ).

6 ! Now, if a quadratic equation is written in general form, y = ax + bx + c, we can rewrite that in Vertex Form by completing the square. However, if we generalize that, we can find a shortcut. Remember, to complete the square, the coefficient of the quadratic term, a, must be.! ax + bx + c = 0 Given x + b a x + c a = 0 Div by a! x + b Rewrite a x + c a + = 0 b a!! x + b a x + b a + c a b a = 0 Add/Subtract (x + b ac b a ) + = 0 a Factor/combine terms Now writing this in two variables, we have:! y = (x + b ac b. a ) + a ac b Replacing! with k, we have! y = (x + b which is now a a ) + k b written in vertex form. The x-coordinate of the vertex is!. To find the a y-coordinate, we substitute that value into the original equation. Example 7 Graph y = x x + b Using!, we have! a ( ) () = + = So the vertex is at (, plug in ) (, 7) Now we can pick a convenient value of x, say 0, find the corresponding y, then use symmetry to find a third point.

7 The benefit we just found is the generalization of completing the square. Rather than completing the square for each problem and getting bogged down in arithmetic, we can now look at an equation that is not in Vertex Form and find the vertex by using b/a, and substituting that back into the original equation. y = x x V(, 7) We now have three ways of graphing a parabola: a) Plotting points (not recommended) b) Using Vertex Form, y = a(x h) + k, V(h, k), pick a convenient value of x, find corresponding y, find 3rd point by symmetry. c) Using General Form, y = ax + bx + c, find vertex by placing b/a, then substituting that value into the equation to find the y-coordinate of the vertex. Find a nd point by picking a convenient value of x and corresponding y, then find a 3rd point by symmetry.