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What are the coordinates of the vertex y equals x plus 2 minus 4?
y = x + 2 - 4 is the same as y = x - 2 which is the equation of a straight line. A single straight line cannot have a vertex.
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)
The vertex of the parabola below is at the point -3 -5 Which of the equations below could be the equation of this parabola?
2
What is the y-coordinate of the vertex of a parabola with the following equation?
The y coordinate is given below:
What is the solution for the equation -x plus y equals 4 and the equation 2x plus y equals 4?
x = 0 and y = 4
What is the vertex of y equals x-3 plus 2?
The equation is linear and so has no vertex.
What is the vertex of y equals x plus 1?
y = x +1 is the equation of a straight line and so has no vertex.
How do you write y equals x minus 4 x plus 2 in vertex form and find the vertex?
The given equation is y = x - 4x + 2 which can be written as y = -3x + 2 This is an equation of a straight line. Therefore it has no vertex and so cannot be written in vertex form.
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 ?
Find the vertex and equation of the directri for y2 equals -32x?
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
What is the equivalent of the following equation y equals x2 - 8x plus 29?
The vertex form is y = (x - 4)2 + 13
What are the coordinates of the vertex y equals x plus 2 minus 4?
y = x + 2 - 4 is the same as y = x - 2 which is the equation of a straight line. A single straight line cannot have a vertex.
What is the equation for vertex form?
The vertex form for a quadratic equation is y=a(x-h)^2+k.
What is the vertex of the equation Y equals 4X squared minus 8Xplus 9?
To find the vertex of the quadratic equation ( Y = 4X^2 - 8X + 9 ), we can use the vertex formula ( X = -\frac{b}{2a} ). Here, ( a = 4 ) and ( b = -8 ), so ( X = -\frac{-8}{2 \times 4} = 1 ). Substituting ( X = 1 ) back into the equation gives ( Y = 4(1)^2 - 8(1) + 9 = 5 ). Therefore, the vertex of the equation is at the point ( (1, 5) ).
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.
Find the equation of the axis symmetry and the coordinates of the vertex of the graph of each function for y equals 2x plus 4?
I'm assuming that you meant y = 2(x^2) +4. If it were only y = 2x +4, then this would be a linear equation and not a parabola. Anyways, use the equation x = -b/2a to find the x-value of your vertex AND your axis of symmetry. (Given the standard equation y = a(x^2) + bx + c) So, x = -0/2(2) - x = 0 (Axis of Symmetry) Now plug 0 back into your equation to find your y-value of your vertex. y = 2(0^2)+4 y=0 + 4 y = 4 Therefore Vertex = (0,4)
What is the y-coordinate of the vertex of the parabola that is given by the equation below?
We will be able to identify the answer if we have the equation. We can only check on the coordinates from the given vertex.
