# What would be the considering function f(x) = x^2 + 4x +1? How would I find the x-coordinate of vertex of this parabola?Once I get this answer I can complete the table to graph my function.

*print*Print*list*Cite

### 3 Answers

Consider this equation:

ax^2+bx=0

x(ax+b)=0

x=0 or ax+b=0

x=0 or ax=-b

x=0 or x=-b/a

So, the two x-intercepts therefore would be 0 and -b/a

Getting the average of the two x-intercepts, which would be the vertext of the curve, then the vertex is:

average= (0-b/a)/2= (-b/a)/2= -b/2a (vertex point)

Now, let's look at the quadratic equation, f(x) (or y)= x^2+4x+1

-b/2a= -4/2(1)= -4/2=** -2 **(x-coordinate)

The standard form of the parabola is:

Y^2= 4aX. Its vertex is at cordinates:(0,0)and focus is at(a,0), Similarly,

**x^2= 4ay** is a parabola, with vertex at (0,0) and focus at (0,a)

Given form is f(x) =x^2+4x+1 .

f(x) =(x+2)^2 - 2^2 +1

f(x)=(x+2)^ -3

f(x)+3=(x+2)^2 or

(x+2)^2= y-3

(x+2)^2= 4(1/4)(y-3) which is of the form,

**X^2= 4aY**.

Therfore, (X,Y) = (0,0) gives :

x+2=0 and y-3=0

x=-2 and y=3.

Therefore, the vertex coordinates are **(x,y)= (-2, 3)**.

The** x **coordinate is** -2**

In order to find out the x coordinate of f(x)'s vertex, you have to know the coordinates of vertex.

If the function is f(x)=ax^2+bx+c, the coordinates of vertex are:

x coordinate= -b/2a

y coordinate = -delta/4a

delta=b^2- 4ac

So, a=1, b=4, c=1 from

f(x) = x^2+4x+1

x coordinate=-4/2*1=-2