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.
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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
So, a=1, b=4, c=1 from
f(x) = x^2+4x+1
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
(x+2)^2= 4(1/4)(y-3) which is of the form,
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
Consider this equation:
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)
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