Easier to show you a simple example as I forget the formulaic approach.
X2 + 4X - 6 = 0
add 6 to each side
x2 + 4X = 6
Now, halve the linear term ( 4 ), square it and add it to both sides
X2 + 4X + 4 = 6 + 4
gather the terms on the right side and factor the left side
(X + 2)2 = 10
subtract 10 from each side
(X + 2)2 - 10 = 0
(- 2, - 10 )
-------------------the vertex of this quadratic function
2 AND 9
Yes, the coordinates for the vertex of a quadratic function in the form (y = ax^2 + bx + c) can be found using the formula (x = -\frac{b}{2a}) to determine the x-coordinate. Once you have the x-coordinate, substitute it back into the original equation to find the corresponding y-coordinate. This gives you the vertex in the form ((x, y)).
To determine the quadratic function from a graph, first identify the shape of the parabola, which can open upwards or downwards. Look for key features such as the vertex, x-intercepts (roots), and y-intercept. The standard form of a quadratic function is ( f(x) = ax^2 + bx + c ), where ( a ) indicates the direction of the opening. By using the vertex and intercepts, you can derive the coefficients to write the specific equation of the quadratic function.
A common technique to rewrite a quadratic function in standard form ( ax^2 + bx + c ) to vertex form ( a(x - h)^2 + k ) is called "completing the square." This involves taking the coefficient of the ( x ) term, dividing it by 2, squaring it, and then adding and subtracting this value inside the function. By rearranging, you can express the quadratic as a perfect square trinomial plus a constant, which directly gives you the vertex coordinates ( (h, k) ).
Do you have a specific vertex fraction? vertex = -b/2a wuadratic = ax^ + bx + c
it is a vertices's form of a function known as Quadratic
It if the max or minimum value.
vertex
2 AND 9
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.
It is a turning point. It lies on the axis of symmetry.
The minimum is the vertex which in this case is 0,0 or the origin. There isn't a maximum.....
vertex
The vertex must be half way between the two x intercepts
The standard form of the quadratic function in (x - b)2 + c, has a vertex of (b, c). Thus, b is the units shifted to the right of the y-axis, and c is the units shifted above the x-axis.
The vertex.
The zeros of a quadratic function, if they exist, are the values of the variable at which the graph crosses the horizontal axis.