Vertex form of a quadratic equation

Transcription

1 Verte form of a quadratic equation Nikos Apostolakis Spring 017 Recall 1. Last time we looked at the graphs of quadratic equations in two variables. The upshot was that the graph of the equation: k = a( h) (1) is a parabola and looks eactl like the graph of = a, but with verte translated at (h,k). For eample here is the graph of the equation +1 = (+) Verte: (, 1) Ais of smmetr: = -intercepts: ( 3, 0), ( 1, 0) -intercept: (0, 3) Smmetric to -intercept: ( 4, 3) The ais of smmetr is the vertical line passing through the verte. In this eample the ais of smmetr is = The -intercepts are obtained b solving the equation we get b setting = 0. In our case: 1 = (+) + = ± 1 + = ±1 + = 1, or + = 1 = 1, or = 3 The -intercept is the point with = 0. Plugging = 0 in the equation we get +1 = () +1 = 4 = 3 To find the point smmetric to the -intercept, we notice that 0 is units to the right of the verte. So the -coordinate of the point is units to the left of the -coordinate of the verte, i.e. at 4. 1

2 Eample 1. Graph the equation: + = ( 1) Answer. The verte is at (1, ), so the ais of smmetr is = 1. To find the -intercepts we put = 0, and solve: = ( 1) 1 = ( 1) This equation has no real solutions, because negative numbers have imaginar square roots. It follows that the parabola has no -intercepts. The -intercept is obtained b setting = 0: + = ( 1) + = = 4 To find the point smmetric to the -intercept, we observe that 0 is 1 unit to the right of the verte, so the smmetric point will be 1 unit to the right of the verte. So the point smmetric to the -intercept is (, 4). In sum the graph is as follows: Now the question is: Question. If we start with an equation in standard form how can we find the verte form (1)? = a +b+c () Answer. We complete the square, of course! Instead of doing it ever time we have an equation in standard form we will do it once for the general form (), and obtain formulas. That s the power Page

3 of algebra! So here goes: = a +b+c c = a ( + ba ) c a c a c a = + b a ( ) b + = + b a + ( + b 4a = + b ) c+ b 4a = a + b 4ac 4a ( + b = a + D 4a = a ( + b where D stands for our old friend, the discriminant. The calculations above show that: ) ( + b ) ) ( ) b Fact 1. The coordinates of the verte of an equation in standard form () are given b: h = b, k = D (3) Eample. Graph the parabola = Accuratel identif the verte, the ais of smmetr, the -intercepts if present, the -intercept, and the point smmetric to the -intercept. Answer. We first find the coordinates of the verte, using the formulas (3): For k we need first to calculate the discriminant: so h = b = 6 = 3 D = b 4ac = ( 1) = 36+4 = 40 k = D 4a = 40 4 = 10 The verte is thus at ( 3, 10). To find the -intercepts we solve the equation: B the quadratic formula the solutions are +6 1 = 0 = b± D = 6± 40 = 6± 10 = 3± 10 Page 3

4 These are irrational numbers. We can use a calculator to get an approimation: , or To find the -intercept we put = 0. We get = 1. 0 is 3 units to the right of the verte, so the point smmetric to the -intercept is 3 units to the left of the verte. So it has coordinates ( 6, 1). In sum, we have the following graph: Eample 3. Graph the parabola =. Accuratel identif the verte, the ais of smmetr, the -intercepts if present, the -intercept, and the point smmetric to the -intercept. Answer. First we find the discriminant: D = b 4ac = ( ) 4(1)( ) = 4+8 = 1 Now we calculate the verte: h = b = = 1 k = D 4a = 1 4 = 3 So the verte is at (1, 3) and the ais of smmetr is = 1. The -intercepts: = b± D = ( )± 1 = ± 3 = 1± 3 Page 4

5 We use the calculator to find approimations to the -intercepts: , The -intercept is =. 0 is 1 unit to the left of the verte so the point smmetric to the -intercept is 1 unit to the right of the verte, so it is (, ). In sum we have the following: Now ou practice: For each of the following equations identif the verte, the ais of smmetr, the -intercepts if present, the -intercept, and the point smmetric to the -intercept. Then sketch a rough graph. 1. = 4+6. = 6 3. = +5+6 The graphs are shown below: = 4+6 = 1 = 6 = 5 = +5+6 Page 5

6 Eercises 1. Graph each of the following quadratic equations: (a) = +4 6 (b) = ++4 (c) = ++6 (d) = 8 (e) = 5+6 (f) = 3 6 (g) = +4. A parabola has ais of smmetr = 3, meets the -ais at the points (1,0) and (5,0), and the -ais at the point (0,10). Find an equation of the parabola in standard form. Page 6