Y=a(x-h)+k is the vertex formula.
Since the vertex is at (-2,-3) this parabola has the
equation:
y=a(x+2)^2-3
We can plug in x=-1 but we really need to know a, to solve for y.
( we can solve it, but we will have an a in the solution)
7
To determine the equation of a parabola with a vertex at the point (5, -3), we can use the vertex form of a parabola's equation: (y = a(x - h)^2 + k), where (h, k) is the vertex. Substituting in the vertex coordinates, we have (y = a(x - 5)^2 - 3). The value of "a" will determine the direction and width of the parabola, but any equation in this form with varying "a" values could represent the parabola.
Interpreting that function as y=x2+2x+1, the graph of this function would be a parabola that opens upward. It would be equivalent to y=(x+1)2. Its vertex would be at (-1,0) and this vertex would be the parabola's only zero.
please help
A parabola that opens upward is a U-shaped curve where the vertex is the lowest point on the graph. It can be represented by the general equation y = ax^2 + bx + c, where a is a positive number. The axis of symmetry is a vertical line passing through the vertex, and the parabola is symmetric with respect to this line. The focus of the parabola lies on the axis of symmetry and is equidistant from the vertex and the directrix, which is a horizontal line parallel to the x-axis.
the vertex of a parabola is the 2 x-intercepts times-ed and then divided by two (if there is only 1 x-intercept then that is the vertex)
5
Parabola: 1+12x-6x^2 Factorizing: -6(x^2 -2x -1/6) Completing the square: -6((x-1)^2 -1 -1/6) => -6(x-1)^2 +7 Vertex of parabola is at: (1, 7)
The vertex of a parabola is the minimum or maximum value of the parabola. To find the maximum/minimum of a parabola complete the square: x² + 4x + 5 = x² + 4x + 4 - 4 + 5 = (x² + 4x + 4) + (-4 + 5) = (x + 2)² + 1 As (x + 2)² is greater than or equal to 0, the minimum value (vertex) occurs when this is zero, ie (x + 2)² = 0 → x + 2 = 0 → x = -2 As (x + 2)² = 0, the minimum value is 0 + 1 = 1. Thus the vertex of the parabola is at (-2, 1).
there can be 0, 1, 2, or infinite intercepts for a parabola 0 intercepts occurs when the parabola does not meet the 2nd line 1 occurs when the parabola intersects a line at the vertex 2 occurs when the line does not intersect at the vertex, but still intersects the parabola infinite occurs when there are 2 parabolas, that although they may be written differently, are the same on a graph.
3
For a parabola with an axis of symmetry parallel to the x-axis, the equation of a parabola is given by: (y - k)² = 4p(x - h) Where the vertex is at (h, k), and the distance between the focus and the vertex is p (which can be calculated as p = x_focus - x_vertex). For the parabola with vertex (1, -3) and focus (2, -3) this gives: h = 1 k = -3 p = 2 - 1 = 1 → parabola is: (y - -3)² = 4×1(x - 1) → (y + 3)² = 4(x - 1) This can be expanded to: 4x = y² + 6y + 13 or x = (1/4)y² + (3/2)y + (13/4)
-2
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.
The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.
The vertex of this parabola is at -2 -3 When the y-value is -2 the x-value is -5. The coefficient of the squared term in the parabola's equation is -3.
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