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.
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 y coordinate is given below:
x = 0 and y = 4
The equation 6 + y = 12 is a simple equation in one variable. If you add 6 to both sides of the equals sign, you get y = 18, and that is the solution.
The equation is linear and so has no vertex.
y = x +1 is the equation of a straight line and so has no 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.
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 ?
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
The vertex form is y = (x - 4)2 + 13
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.
The vertex form for a quadratic equation is y=a(x-h)^2+k.
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.
We will be able to identify the answer if we have the equation. We can only check on the coordinates from the given vertex.
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)
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).