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)
-2
To find other points on a parabola, you can use its equation, typically in the form (y = ax^2 + bx + c). By selecting different values for (x) and substituting them into the equation, you can calculate the corresponding (y) values. Alternatively, you can also use the vertex form, (y = a(x-h)^2 + k), where ((h, k)) is the vertex, to find points by choosing (x) values around the vertex. Plotting these points will help visualize the shape of the parabola.
2
Use this form: y= a(x-h)² + k ; plug in the x and y coordinates of the vertex into (h,k) and then the other point coordinates into (x,y) and solve for a.
3
The vertex form for a quadratic equation is y=a(x-h)^2+k.
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}))).
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.
-2
In the equation y x-5 2 plus 16 the standard form of the equation is 13. You find the answer to this by finding the value of X.
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.
The difference between standard form and vertex form is the standard form gives the coefficients(a,b,c) of the different powers of x. The vertex form gives the vertex 9hk) of the parabola as part of the equation.
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)
please help
y=2(x-3)+1
Y=3x^2 and this is in standard form. The vertex form of a prabola is y= a(x-h)2+k The vertex is at (0,0) so we have y=a(x)^2 it goes throug (2,12) so 12=a(2^2)=4a and a=3. Now the parabola is y=3x^2. Check this: It has vertex at (0,0) and the point (2,12) is on the parabola since 12=3x2^2