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What is the vertex of y -x2 4x -6?

To find the vertex of the quadratic function ( y = -x^2 + 4x - 6 ), we can use the vertex formula ( x = -\frac{b}{2a} ). Here, ( a = -1 ) and ( b = 4 ), so ( x = -\frac{4}{2(-1)} = 2 ). Substituting ( x = 2 ) back into the equation gives ( y = -(2)^2 + 4(2) - 6 = 2 ). Therefore, the vertex is at the point ( (2, 2) ).


What is the range of y equals 2x2 plus 4x-3?

y = 2x2 + 4x - 3 This equation describes a parabola. Because it's first term is positive, we know that it goes infinitely upward, and has a minimum that occurs at it's vertex. You can find it's vertex by taking it's derivative and solving for zero: y' = 4x + 4 0 = 4x + 4 0 = x + 1 x = -1 y = (-1)2 + 4(-1) - 3 y = 1 - 4 - 3 y = -6 So the vertex is at (-1, -6), which means that y ≥ -6


How do you calculate the vertex on a function graph?

Example function.Y = X2 - 4X + 5set to 0X2 - 4X + 5 = 0X2 - 4X = - 5Now, halve the linear coefficient, square it and add it to both sidesX2 - 4X + 4 = - 5 + 4gather terms on the right and factor on the left(X - 2)2 = -1==============Vertex form.(2, - 1)=======Vertex.


What are the coordinates of the vertex of y x2 - 4x - 5?

Y = X2 - 4X - 5set to zeroX2 - 4X - 5 = 0X2 - 4X = 5halve the linear term ( - 4 ) then square it and add that result to both sidesX2 - 4X + 4 = 5 + 4factor on the left and gather terms together on the right(X - 2)2 = 9(X - 2)2 - 9 = 0==============vertex form(2, - 9)======vertex


What is the vertex of y-2x2-4x-5?

2x2= 4


What is the x-coordinate of the vertex of y equals 4x - 12 - 3?

y=4x-12-3 is the equation of a straight line. It does not have a vertex. Did you mean y=x squared - 12x - 3 ?


What is the vertex of the parabola y-2x2-4x 4?

Assuming the missing symbol there is an equals sign, then we have: y - 2x2 - 4x = 4 We can find it's vertex very easily by solving for y, and finding where it's derivative equals zero: y = 2x2 + 4x + 4 y' = 4x + 4 0 = 4x + 4 x = -1 So the vertex occurs Where x = -1. Now we can plug that back into the original equation to find y: y = 2x2 + 4x + 4 y = 2 - 4 + 4 y = 2 So the vertex is at the point (-1, 2)


What is the vertex form of y equals x2 plus 4x - 7?

The vertex of the positive parabola turns at point (-2, -11)


How do you find a vertex in quadratic function?

Easier to show you a simple example as I forget the formulaic approach. X2 + 4X - 6 = 0 add 6 to each side x2 + 4X = 6 Now, halve the linear term ( 4 ), square it and add it to both sides X2 + 4X + 4 = 6 + 4 gather the terms on the right side and factor the left side (X + 2)2 = 10 subtract 10 from each side (X + 2)2 - 10 = 0 (- 2, - 10 ) -------------------the vertex of this quadratic function


What is the vertex of the quadratic y -2x2 plus 4x-1?

It is (1, 1).


Find the x intercept or interceptss and the coordinates of the vertex for the parabola x2-4x-12 if there is more than one x intercept separate them with commas?

-2, 6


What is the equation of a parabola with the vertex of 2 -1?

3

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