Since -(-2)=2 and carrying over the coefficient a=2 we converted to Vertex Form
$$ y=2(x+2)²-1 $$
We are given the following Quadratic Equation in Standard Form $$ y=3x²- 6x-2 $$ .
First, compute the x-coordinate of the vertex
$$ h = – b/2a = -(-6) / (2*3) = 1 $$ .
Next, compute the y-coordinate of the vertex by plugging h=1 into the given equation:
$$ k = 3*(1)²-6(1)-2 = -5 $$ .
Therefore, the vertex is
$$ (h,k)=(1,-5) $$ .
Thus, we transformed the above Standard Form into the Vertex Form $$ y=3(x-1)²-5 $$ Easy, wasn’t it?
Tip: When using the above Standard Form to Vertex Form Calculator to solve
$$ 3x²-6x-2=0$$
we must enter the 3 coefficients a,b,c as
$$ a=3, b=-6, c=-2 $$.
Then, the calculator will find the Vertex $$ (h,k)=(1,-5) $$ Step by Step.
Finally, the Vertex Form of the above Quadratic Equation is $$ y=(x-1)²-5 $$ Get it now? Try the our Standard Form to Vertex Standard Calculator again.
There are two ways to find h and k, the vertex x- and y- coordinates. There is a fast way and a long way.
1) The fast way: Given $$ y = ax²+bx+c $$ we first compute $$ h = -b / 2a $$
and next $$ k=f(h) $$ .
Example: $$ y=3x²+6x+4 $$ thus $$ h =-6/2*3 = -1 $$ and
$$k = f(-1) = 3(-1)²+6(-1)+4 = 3-6+4 = 1 $$
Thus, Vertex Coordinates are $$ (k,h)=(-1,1) $$.
2) The long way: We do the Complete-the-Square procedure to convert
$$ y=ax²+bx+c $$ into
$$ y=a(x-h)²+k $$.
We created a separate page to learn this method.
Complete the Square here .
h and k are the Vertex x- and y- coordinates of the Graph of a Quadratic Equation. They give the Location of a Minimum (when a>0) or Maximum (when a<0).
You may also think of h and k as shifts/transformations:
Shifting the Standard Parabola
$$ y=x² $$
h units right yields
$$y=(x-h)² $$.
Shifting it k units up yields
$$y=(x-h)²+k $$ .
By performing those 2 shifts we moved the Vertex from
the origin (0,0) to the new location $$ (h,k) $$ .
