Question #6d616

1 Answer
Apr 9, 2016

Answer:

Y intercept: At the point #(0,1)#
Axis of symmetry: At #x=1/2#
Vertex: At the point #(1/2,0)#

Explanation:

Let's have a look at the y-intercept first:

Y intercepts where x =0.

Therefore, substitute all x variables with 0 and solve for "y".

#y=4(0)^2-4(0)+1#
#y=1#

Therefore, the y intercept is present at the point (0,1)

Now let's have a look at the vertex:

Note: I'll use the terms Turning Point and Vertex interchangeably as they are the same thing.

Consider the general form of a parabolic function:
#y=ax^2+bx+c#
If we compare the equation that you have presented:

#y = 4x^2 -4x +1#

We can determine that:

The #x^2# coefficient is 4; this implies that #a# = 4
The #x# coefficient is -4; this implies that #b# = -4
The constant term is 1; this implies that #c# = 1

Therefore, we can use the formula:

#Tp_x=-b/(2a)# to determine the #x# value of the turning point.

Substituting the appropriate values into the formula we get:

#Tp_x=4/(2*4)#

#=4/8#

#=1/2#

Therefore, the #x# value of the vertex is #x = 1/2#

Substitute #x = 1/2# into the general form of the equation to determine the coordinate of the vertex:

#y = 4x^2 -4x +1#
Therefore, #y=4(1/2)^2-4(1/2)+1#
Simplifying the function we get:

#y=0#

Therefore, the vertex of the function is present at the point #(1/2,0)#

Finally, let's have a look at the axis of symmetry:

The axis of symmetry is essentially the #x# value of the turning point (the vertex) of a parabola.

If we have determined the #x# value of the turning point as #1/2#, we can then say that the axis of symmetry of the function is present at #x=1/2#.