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Title: The standard form of a parabola is x=y^2+6y+1 What is the vertex form of the equation? Post by: rkoch on Nov 4, 2012 What is the vertex form of the equation
Title: The standard form of a parabola is x=y^2+6y+1 What is the vertex form of the equation? Post by: IMme on Nov 4, 2012 The vertex formula is (-b)/2a. Therefore, you would replace b with 6 and a with 1. Keeping that in mind, we get: (-6)/2. By dividing we see that 2 times -3 is -6, so therefore the vertex is at -3.
Title: The standard form of a parabola is x=y^2+6y+1 What is the vertex form of the equation? Post by: roarie on Nov 4, 2012 All you have to do is use the vertex formula: -b/2a.
1.) Plug in the -b value. Remember that standard form of a quadratic equation is ax^2+bx+c. So for -b we'll plug in 6. so -6/2a. 2.) Plug in the a value. Here a would be y, which is also know as 1. 3.) Solve the expression. -6/2(1). 4.) The vertex would be -3. Title: The standard form of a parabola is x=y^2+6y+1 What is the vertex form of the equation? Post by: $abood$ on Nov 4, 2012 Like the first guy said complete the square. x = y^2 + 6y + 9 - 9 + 1 x = (y + 3)^2 - 8
x + 8 = (y + 3)^2 vertex (-8, -3) Title: The standard form of a parabola is x=y^2+6y+1 What is the vertex form of the equation? Post by: smokinjoe531 on Nov 4, 2012 x = y^2 + 6y + 1
x = y^2 + 6y + 9 - 8 x = (y + 3)^2 - 8 (x + 8) = (y + 3)^2 vertex coordinate = (-8,-3) |