0
3
What else can I help you with?
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 y equals negative 2 point 5 parenthese x minus 4 parenthese end squared minus 5?
y = -5 By using calculus, the derivative of y = -2.5(x-4)2 - 5 is y' = -5(x-4). Solving the equation -5(x-4) = 0 gives x = 4 (since the slope of the parabola at the vertex is zero). Plug this back into the equation: y = -2.5(4 - 4) -5 = -5, so the y-coordinate is -5. The equation of the parabola is given in the vertex form y = a(x - h)2 + k, where (h, k) is the vertex. So the vertex is (4, -5).
