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.
x = 4 and y = 7
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).
x = 4 and y = 7 which will satisfy both equations
An equation where x equals 3.
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.
(1/2, 71 and 3/4)or(0.5, 71.75)
The vertex is (5, 11).
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.
Question can be taken as multiple meanings. Please see discussion.
The required equation is: -7x = 63
its a simple parobola symmetric about y axis, having its vertex at (0,-4). we can make its graph by changing its equation in standard form so that we can get its different standard points like vertex, focus, etc.
x = 4 and y = 7
Number of sides minus two equals number of diagonals drawn from one vertex.
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).
x = 4 and y = 7 which will satisfy both equations
An equation where x equals 3.