Do you have a specific vertex fraction?
vertex = -b/2a
wuadratic = ax^ + bx + c
y=2(x-3)+1
2
2 AND 9
You would convert it to vertex form by completing the square. You can also find the optimum value as optimum value and vertex are the same.
To find the vertex of a quadratic equation in standard form, (y = ax^2 + bx + c), you can use the vertex formula. The x-coordinate of the vertex is given by (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. The vertex is then the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))).
it is a vertices's form of a function known as Quadratic
The vertex form for a quadratic equation is y=a(x-h)^2+k.
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.
y=2(x-3)+1
2
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)
It if the max or minimum value.
The vertex.
vertex
y= -5/49(x-9)^2+5
2 AND 9
You should always use the vertex and at least two points to graph each quadratic equation. A good choice for two points are the intercepts of the quadratic equation.