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When dealing with parametric equations, each variable, often denoted as \( x \) and \( y \), is expressed in terms of a third variable such as \( t \). This allows for a more flexible representation of curves. However, converting these into rectangular form combines the \( x \) and \( y \) variables into a single equation without the parameter \( t \).
The given parametric equations are:

• \( x = 4t + 3 \)
• \( y = 16t^2 - 9 \)

To convert these into rectangular form, you need to eliminate \( t \) and relate \( x \) directly to \( y \). This is done by expressing \( t \) in terms of \( x \) from the first equation, \( t = \frac{x-3}{4} \), and substituting back into the second equation. Through careful simplification, you end up with a single equation for \( y \) in terms of \( x \), like \( y = x^2 - 6x \).
Thus, converting parametric to rectangular form is all about removing the parameter and defining the relationship between \( x \) and \( y \).

The domain of a function is a set of all possible input values, typically \( x \), that allow the function to work without issues like division by zero or taking square roots of negative numbers.
In rectangular form \( y = x^2 - 6x \), the expression is a polynomial, which is defined for all real numbers. Therefore, there are no restrictions on \( x \), and the domain is all real numbers, denoted as \( (-\infty, \infty) \).
Knowing the domain is essential because it tells you where the function exists and where it can be graphed without any undefined behavior.

To solve for \( t \) in a parametric equation, you typically take the equation for \( x \) or \( y \) that is simplest and solve it algebraically. Here, starting with \( x = 4t + 3 \), the goal is to isolate \( t \).

• Subtract 3 from both sides to get: \( x - 3 = 4t \)
• Then, divide each side by 4: \( t = \frac{x-3}{4} \)

After finding \( t \) in terms of \( x \), you can substitute this expression into the other equation. This allows you to transform the system of parametric equations into a single rectangular equation. Solving for \( t \) is a critical step in finding the rectangular form.

Graphing functions involves plotting points on a coordinate plane to visualize the behavior of an equation. Once you have the rectangular form, \( y = x^2 - 6x \), you can graph it to see the shape and intersection points.
For polynomials like \( y = x^2 - 6x \), the graph is a parabola. It opens upwards since the \( x^2 \) term is positive. The vertex and intercepts give insights into the curve's position:

• The vertex can be found using \( x = -\frac{b}{2a} \) where the equation is \( ax^2 + bx + c \)
• For \( x^2 - 6x \), \( a = 1 \) and \( b = -6 \), giving the vertex at \( x = 3 \)
• Plugging back, \( y = 9 - 18 = -9 \)
• Intercepts are found by setting \( y \) to 0 or using the factored form \((x-3)^2 = 9\)

Graphing these points and connecting them with a smooth curve will fully depict the function. Understanding graph shapes helps in interpreting equations and their real-world implications.