# How do you find the solution to the parabola for the quadratic formula?

 AnonymousFrom: -Posts: -Votes: - I dont get how would you find the solutions or points that the parabola crosses the x axis. ex- Graph the quadratic equation 2x2 – 4x – 6 = 0 on your own graph paper. Choose the solution or solutions below that correspond to your graph. You may need to select more than one. x = -1 x = -3 x = 3 x = 1 No Solution I thought it was 1 and -3 but apperently i was wrong. Please help me.

 Bitty PottyFrom: USPosts: 37Votes: 74 I assume when you say 2x2 – 4x – 6 = 0 you mean 2x^2 – 4x – 6 = 0 Technically, that's not a parabola. But you can graph y = 2x^2 – 4x – 6, which is a parabola. First, it might help to find the vertex. One way to do that is by completing the square, i.e. if you can get the equation into the form y = a(x - b)^2 - c, then the vertex is (b,c)? y = 2(x^2 - 2x - 3) y = 2(x^2 - 2x + 1 - 4) y = 2(x^2 - 2x + 1) - 8 y = 2(x-1)^2 + 8 So the vertex is (1,-8) Another way to find the vertex is to find dy/dx: dy/dx = 4x - 4 Now where is dy/dx = 0? 0 = 4x - 4 4x = 4 x = 4/4 = 1 If x = 1, then y = 2(1)^2 - 4(1) - 6 = -8 So the vertex is (1,-8) Now that you know the vertex, choose two other points to graph, for example where the graph crosses the x-axis (i.e. where y=0). So we have: 2x^2 – 4x – 6 = 0 You can factor that: 2(x^2 – 2x – 3) = 0 2(x+1)(x-3) = 0 So either x +1 = 0 or x - 3 = 0 So either x = -1 or x = 3 Or, you can use the Quadratic Formula: x = (-(-4) +/- sqrt((-4)^2 - 4(2)(-6))) / 2(2) x = (4 +/- sqrt(16 + 48)) / 4 x = (4 +/- sqrt(64)) / 4 x = (4 +/- 8) / 4 x = (12 or -4) / 4 x = 3 or -1 So you can now plot 3 points on the graph: (1,-8), (3,0), and (-1,0) If you want, you can plot a few more points to help you see what the graph should look like. Then you can fill in the rest of the parabola with a freehand approximation. To see exactly what the graph should look like, try http://www.walterzorn.com/grapher/graphe…

 GamaHeadFrom: USPosts: 7Votes: 21 If you plug in x = 1, you should get this: = 2(1)^2 - 4(1) - 6 = 2 - 4 - 6 = -2 - 6 = -8 That is not zero, so x = 1 is not a solution. If you plug in x = -3, you should get this: = 2(-3)^2 - 4(-3) - 6 = 2(9) +12 - 6 = 18 + 12 - 6 = 30 - 6 = 24 This is not zero, so x = -3 is not a solution. Just keep trying with the other values.

 AnonymousFrom: -Posts: -Votes: - Factor out 2 2(x^2 -2x -3) = 0 2(x-3)(x+1) = 0 x = 3 or x = -1 Your error is the fact that you just took the zero at face value Think of it this way, x-3 = 0 x = 3 That way you won't get confused.