Question #6d616

Apr 9, 2016

Y intercept: At the point $\left(0 , 1\right)$
Axis of symmetry: At $x = \frac{1}{2}$
Vertex: At the point $\left(\frac{1}{2} , 0\right)$

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".

$y = 4 {\left(0\right)}^{2} - 4 \left(0\right) + 1$
$y = 1$

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:
$y = a {x}^{2} + b x + c$
If we compare the equation that you have presented:

$y = 4 {x}^{2} - 4 x + 1$

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:

$T {p}_{x} = - \frac{b}{2 a}$ to determine the $x$ value of the turning point.

Substituting the appropriate values into the formula we get:

$T {p}_{x} = \frac{4}{2 \cdot 4}$

$= \frac{4}{8}$

$= \frac{1}{2}$

Therefore, the $x$ value of the vertex is $x = \frac{1}{2}$

Substitute $x = \frac{1}{2}$ into the general form of the equation to determine the coordinate of the vertex:

$y = 4 {x}^{2} - 4 x + 1$
Therefore, $y = 4 {\left(\frac{1}{2}\right)}^{2} - 4 \left(\frac{1}{2}\right) + 1$
Simplifying the function we get:

$y = 0$

Therefore, the vertex of the function is present at the point $\left(\frac{1}{2} , 0\right)$

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 $\frac{1}{2}$, we can then say that the axis of symmetry of the function is present at $x = \frac{1}{2}$.