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ViviY = X^2 - X - 23
set to 0
X^2 - X - 23 = 0
X^2 - X = 23
now, complete the square by halving the linear term ( - 1 ), and squaring it, then add it to both sides
X^2 - X + 1/4 = 23 + 1/4
factor the left side; gather terms right
(X - 1/2)^2 = 93/4
(X - 1/2)^2 - 93/4 = 0
(1/2,93/4)
---------------------Vertex
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Where is the vertex coordinate of the parabola y equals x square -3x -10 located on the Cartesian plane?
7
What is the y-coordinate of the vertex of a parabola with the following equation y equals x2 - 8x plus 18?
You can work this out by taking the derivative of the equation, and solving for zero: y = x2 - 8x + 18 y' = 2x - 8 0 = 2x - 8 x = 4 So the vertex occurs where x is equal to 4. You can then plug that back into the original equation to get the y-coordinate: y = 42 - 8(4) + 18 y = 16 - 32 + 18 y = 2 So the vertex of the parabola occurs at the point (4, 2), leaving 2 as the answer to your question.
What is the vertex of the function y equals 2 x 2 plus 8x plus 1?
If you meant, your two equations are y=2x+2 and y= 8x+1 your vertex will be at x=1/6 and y=7/3 you can plug these in and they will check out. but if you meant, y=(2x2) +8x+1 y=4+8x+1 y=8x+5 and then you only have one line so there will be no vertex. but the y intercept is at (0, 5) and your x-intercept at (-5/8, 0) if neither are these are what you are saying, sorry, and maybe try and repost and I will help you out.
How do you find the axis of symmetry and vertex of y equals x squared plus 6x plus 10?
By completing the square y = (x+3)2+1 Axis of symmetry and vertex: x = -3 and (-3, 1) Note that the parabola has no x intercepts because the discriminant is less than zero
What is the x-coordinate of the of the vertex of y equals -3x2 plus 12x-5?
You can find the x-coordinate of it's vertex by taking it's derivative and solving for zero: y = -3x2 + 12x - 5 y' = -6x + 12 0 = -6x + 12 6x = 12 x = 2 Now that we have it's x coordinate, we can plug it back into the original equation to find it's y coordinate: y = -3x2 + 12x - 5 y = -3(2)2 + 12(2) + 5 y = -12 + 24 + 5 y = 17 So the vertex of the parabola y = -3x2 + 12x - 5 occurs at the point (2, 17).
