If you mean: y = x^2 -6x +6 then the vertex is at (3, -3)
If you mean: y = x^2 -6x +6 then the vertex is at (3, -3)
y=x2-6x+9
4
y = - x2 +6x - 5.5
Y = X2 + 6X + 2set to 0X2 + 6X + 2 = 0X2 + 6X = - 2now, halve the linear coefficient ( 6 ), square it and add it to both sidesX2 + 6X + 9 = - 2 + 9gather terms on the right and factor the left(X + 3)2 = 7(X + 3)2 - 7 = 0==============vertex form(- 3, - 7)=======vertex
The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.
помогите плиз сократить дробь 3y^2-8y-6/4-9y^2
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The question does not contain an equation: only an expression. An expression cannot have a vertex form.
(3, -21)
To complete the square for the expression (y = x^2 - 6x + 15), we first rewrite it as (y = (x^2 - 6x) + 15). Next, we complete the square on (x^2 - 6x) by taking half of (-6) (which is (-3)), squaring it to get (9), and rewriting the expression as (y = (x - 3)^2 + 6). The vertex form is (y = (x - 3)^2 + 6), with the vertex at ((3, 6)). Since the coefficient of the squared term is positive, the vertex represents a minimum point.
x = -3y = -14