Consider the Quadratic Function $$ y=(x-2)^{2}-4 $$ N.B. Give all coordinates as ordered pairs. a. Coordinates of vertex: b. Coordinates of y-intercept: c. Domain: d. Range: e. Equation of Axis of Symmetry:

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**a. Coordinates of vertex:**

The function is given by  
$$
y = (x-2)^2 - 4,
$$  
which is in the form  
$$
y = (x-h)^2 + k,
$$  
where $$h = 2$$ and $$k = -4$$. Therefore, the vertex is  
$$
(2, -4).
$$

**b. Coordinates of y-intercept:**

The y-intercept occurs when $$x=0$$. Substitute $$x = 0$$ into the equation:  
$$
y = (0-2)^2 - 4 = 4 - 4 = 0.
$$  
Thus, the y-intercept is  
$$
(0, 0).
$$

**c. Domain:**

The quadratic function is defined for all real values of $$x$$. Therefore, the domain is  
$$
\{ x \in \mathbb{R} \} \quad \text{or} \quad (-\infty, \infty).
$$

**d. Range:**

Since the parabola opens upward and the vertex represents the minimum value, the smallest $$y$$-value is $$-4$$. Thus, the range is  
$$
\{ y \in \mathbb{R} \mid y \geq -4 \} \quad \text{or} \quad [-4, \infty).
$$

**e. Equation of Axis of Symmetry:**

The axis of symmetry is the vertical line that passes through the vertex. Since the x-coordinate of the vertex is $$2$$, the axis of symmetry is  
$$
x = 2.
$$
**a. Vertex:** (2, -4) **b. Y-intercept:** (0, 0) **c. Domain:** All real numbers **d. Range:** $$ y \geq -4 $$ **e. Axis of Symmetry:** $$ x = 2 $$

Consider the Quadratic Function \( y=(x-2)^{2}-4 \) N.B. Give all coordinates as ordered pairs. a. Coordinates of vertex: b. Coordinates of y-intercept: c. Domain: d. Range: e. Equation of Axis of Symmetry:

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