Problem 62 - Elementary Algebra

Question

Given the following quadratic functions, determine the domain and range. $$ f(x)=3 x 2+30 x+50 $$

Step-by-Step Solution

Answer

Domain is \((-\infty, \infty)\); Range is \([-25, \infty)\).

Step 1: Identify the type of function

The function \( f(x) = 3x^2 + 30x + 50 \) is a quadratic function. Quadratic functions are parabolas and can open upwards or downwards depending on the leading coefficient.

Step 2: Determine the domain of the function

For any quadratic function, the domain is all real numbers because you can substitute any real number for \( x \) and get a valid output for \( f(x) \). So, the domain of \( f(x) = 3x^2 + 30x + 50 \) is \( (-\infty, \infty) \).

Step 3: Calculate the vertex of the parabola

The vertex of a parabola given by \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). For \( 3x^2 + 30x + 50 \), \( a = 3 \) and \( b = 30 \). Thus \( x = -\frac{30}{2 \times 3} = -5 \).

Step 4: Determine the y-coordinate of the vertex

Substitute \( x = -5 \) back into the function to find \( y \). So \( f(-5) = 3(-5)^2 + 30(-5) + 50 = 3(25) - 150 + 50 = 75 - 150 + 50 = -25 \). Thus, the vertex is \((-5, -25)\).

Step 5: Establish the range of the function

Since the leading coefficient (3) is positive, the parabola opens upwards. Hence, the minimum point is the vertex. Therefore, the range of the function is \([-25, \infty)\).

Key Concepts

Domain and Range

When dealing with quadratic functions, a key concept is their domain and range. The domain of a quadratic function is quite straightforward: it includes all real numbers. This is because you can plug any real value into the function for the variable

• For the equation \( f(x) = 3x^2 + 30x + 50 \), the domain is
  \( (-\infty, \infty) \), meaning any real number can be used for \( x \).

The range, on the other hand, depends on the orientation and vertex of the parabola. Since this quadratic function opens upwards, due to its positive leading coefficient, the range starts from the vertex's y-value and extends to infinity.

• Here the range is from the y-coordinate of the vertex, which is
  \(-25\), and goes to infinity, noted as
  \([-25, \infty)\).

Vertex of a Parabola

The vertex of a parabola is its peak if it opens downwards or its lowest point when it opens upwards. It's a crucial part of understanding the shape and position of the parabola. In a quadratic formula \( ax^2 + bx + c \), the vertex can be found mathematically using specific formulas for its coordinates.

• The x-coordinate of the vertex is found using \( x = -\frac{b}{2a} \).
  For the function \( f(x) = 3x^2 + 30x + 50 \), the values \( a = 3 \) and \( b = 30 \) lead us to \( x = -5 \).
• To find the y-coordinate, substitute \( x = -5 \) back into the function,
  giving \( f(-5) = -25 \).

Thus, the vertex of this specific parabola is at
\((-5, -25)\). This point helps determine the direction and range.

Parabola Opening Direction

Understanding the direction in which a parabola opens is fundamental in defining its range and interpreting its graph. The sign of the leading coefficient \( a \) in the quadratic function \( ax^2 + bx + c \) dictates whether the parabola opens upwards or downwards. A positive \( a \) means it opens upwards, while a negative \( a \) indicates a downward opening.

• For \( f(x) = 3x^2 + 30x + 50 \), since the leading coefficient
  \( a = 3 \) is positive, the parabola opens upwards.

This means that the vertex represents the minimum point,
providing us with the starting point of the range.
Understanding this opening direction helps in visualizing the
curve and predicting its behavior for different \( x \) values.