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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.
It depends on the level of your mathematical knowledge. One way is to differentiate the quadratic equation and find the value of x for which the derivative is 0. The advantage of this method is that it works for turning points of polynomials of all degrees. The disadvantage is that you need to know differentiation. For a quadratic, an alternative, and simpler way is to write the equation in the form: y = ax2 + bx + c Then the x value of the vertex is -b/2a
3
The vertex form of a quatdratic equation (otherwise called the graphing form) is y=a(x-h)2+k For those of you who don't know what 'h', 'a', and 'k' are, they are parameters. The negative sign in front of the 'h' refers to the opposite of the x coordinate in the vertex. The 'k' refers to the y coordinate in the vertex. 'A' refers to the stretch or compression factor. So, for example, say you have a parabola with a stretch factor of 2 whose vertex coordinates are (-3,4). The equation would be y=2(x+3)2+4 Of course, if a parabola has no stretch/compression factor, there would be no 'a' in the equation. I hope this helped, and good luck!
The vertex form for a quadratic equation is y=a(x-h)^2+k.
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
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0