Question

Which function in vertex form is equivalent to f(x) = x2 + x +1?
O f(x) = (x + 3+
1
3
o f(x)
4
f(x) = (x +
f(x) = (x + +
f(x) = (x + 3D + 1
IN​

Answer 1

Answer:

$\displaystyle f(x)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$

Step-by-step explanation:

We are given the function:

$f(x)=x^2+x+1$.

And we want to turn this into vertex form.

Note that our given function is in the standard form:

$f(x)=ax^2+bx+c$

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:

$f(x)=a(x-h)^2+k$

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:

$\displaystyle x=-\frac{b}{2a}$

Therefore, our point is:

$\displaystyle x=-\frac{(1)}{2(1)}=-\frac{1}{2}$

To find the y-coordinate or k, substitute this value back into the function:

$\displaystyle f\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)+1=\frac{3}{4}$

Thus, the vertex is (-1/2, 3/4). So, h = -1/2 and k = 3/4.

Hence, the vertex form is:

$\displaystyle f(x)=(1)\left(x-\left(-\frac{1}{2}\right)\right)^2+\left(\frac{3}{4}\right)$

Simplify:

$\displaystyle f(x)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$