![\displaystyle f(x)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}](/tpl/images/0708/3957/883ca.png)
Step-by-step explanation:
We are given the function:
.
And we want to turn this into vertex form.
Note that our given function is in the standard form:
![f(x)=ax^2+bx+c](/tpl/images/0708/3957/a2b43.png)
In other words, a = 1, b = 1, and c = 1.
To convert from standard form to vertex form, we can either: (1) complete the square, or (2) find the vertex manually.
In most cases, the second method is more time efficient.
Vertex form is given by:
![f(x)=a(x-h)^2+k](/tpl/images/0708/3957/835d7.png)
Where a is the leading coefficient and (h, k) is the vertex.
We have already determined that a = 1.
Find the vertex. The x-coordinate of the vertex of a quadratic is given by:
![\displaystyle x=-\frac{b}{2a}](/tpl/images/0708/3957/84022.png)
Therefore, our point is:
![\displaystyle x=-\frac{(1)}{2(1)}=-\frac{1}{2}](/tpl/images/0708/3957/09df8.png)
To find the y-coordinate or k, substitute this value back into the function:
![\displaystyle f\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)+1=\frac{3}{4}](/tpl/images/0708/3957/f9ff3.png)
Thus, the vertex is (-1/2, 3/4). So, h = -1/2 and k = 3/4.
Hence, the vertex form is:
![\displaystyle f(x)=(1)\left(x-\left(-\frac{1}{2}\right)\right)^2+\left(\frac{3}{4}\right)](/tpl/images/0708/3957/287d7.png)
Simplify:
![\displaystyle f(x)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}](/tpl/images/0708/3957/883ca.png)