Y = X2 + 6X + 2
set to 0
X2 + 6X + 2 = 0
X2 + 6X = - 2
now, halve the linear coefficient ( 6 ), square it and add it to both sides
X2 + 6X + 9 = - 2 + 9
gather terms on the right and factor the left
(X + 3)2 = 7
(X + 3)2 - 7 = 0
==============vertex form
(- 3, - 7)
=======vertex
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (-2, -3), and a point on it is (-1, -5) → -5 = a(-1 - -2)² + -3 → -5 = a(1)² - 3 → -5 = a - 3 → a = -2 → The coefficient of the x² term is -2.
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (2, -1), and a point on it is (5, 0) → 0 = a(5 - 2)² + -1 → 0 = a(3)² -1 → 1 = 9a → a = 1/9 → The coefficient of the x² term is 1/9
-2
Vertex = (3, - 2)Put in vertex form.(X - 3)2 + 2X2 - 6X + 9 + 2 = 0X2 - 6X + 11 = 0=============The coefficeint of the squared term is 1. My TI-84 confirms the (4, 3) intercept of the parabola and the 11 Y intercept shown by the function.
look for the interceptions add these and divide it by 2 (that's the x vertex) for the yvertex you just have to fill in the x(vertex) however you can also use the formula -(b/2a)
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.
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.
The vertex form for a quadratic equation is y=a(x-h)^2+k.
No, not if the y is squared. When graphed the equation will not form a straight line.
it will form a parabola on the graph with the vertex at point (0,0) and points at (1,1), (-1,1), (2,4), (-2,4)......
To find the coefficient of the squared term in the parabola's equation, we can use the vertex form of a parabola, which is (y = a(x - h)^2 + k), where ((h, k)) is the vertex. Given the vertex at (3, 1), the equation starts as (y = a(x - 3)^2 + 1). Since the parabola passes through the point (4, 0), we can substitute these values into the equation: (0 = a(4 - 3)^2 + 1), resulting in (0 = a(1) + 1). Solving for (a), we find (a = -1). Thus, the coefficient of the squared term is (-1).
The equation you gave, 2y+5 = (x-4)2 can also be expressed as y = 0.5(x-4)2 - 5. In the form a(x-p)2 + q, the vertex is the point (p, q). Thus, the vertex of 2y + 5 = (x-4)2 is (4, -5).
The vertex form is y = (x - 4)2 + 13
To find the coefficient of the squared expression in the parabola's equation, we can use the vertex form of a parabola, which is ( y = a(x - h)^2 + k ), where ((h, k)) is the vertex. Given the vertex is ((-4, -1)), the equation becomes ( y = a(x + 4)^2 - 1 ). When (y = 0) and (x = 2), substituting these values gives (0 = a(2 + 4)^2 - 1), leading to (0 = a(6^2) - 1) or (1 = 36a). Therefore, (a = \frac{1}{36}), which is the coefficient of the squared expression.
To find the coefficient of the squared term in the parabola's equation, we can use the vertex form of a parabola, which is (y = a(x - h)^2 + k), where ((h, k)) is the vertex. Here, the vertex is ((-3, -1)), so the equation becomes (y = a(x + 3)^2 - 1). Given that when (y = 0), (x = 4), we can substitute these values into the equation to find (a): [0 = a(4 + 3)^2 - 1 \implies 0 = a(7^2) - 1 \implies 1 = 49a \implies a = \frac{1}{49}.] Thus, the coefficient of the squared term is (\frac{1}{49}).
The question does not contain an equation: only an expression. An expression cannot have a vertex form.
The graph of a quadratic function is always a parabola. If you put the equation (or function) into vertex form, you can read off the coordinates of the vertex, and you know the shape and orientation (up/down) of the parabola.