Question

Question #6d616

Answer 1

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`.