Problem 49 - College Algebra

Question

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$y^{2}+6 y-x+5=0$$

Step-by-Step Solution

Answer

The vertex of the parabola is \((-3/2, -21/4)\). The parabola opens to the right. The domain of the parabola is \([-21/4, +\infty)\) and the range is \((-\infty, +\infty)\). The relation given by the equation is not a function.

Step 1: Rearrange the equation

Rearrange the equation as follows: \(x=y^{2}+6y-5\), which can be recognized as the standard form \((y-k)^{2}=4p(x-h)\) of the equation of a parabola oriented in a horizontal direction.

Step 2: Complete the square

Complete the square on \(y\). To do so, add \(9/4\) to both sides of the equation to get \(x+5+(3/2)^2=y^{2}+6y+(3/2)^2\). After simplifying, this reads as \(x+41/4=(y+3/2)^2\). Finally, convert back to standard form to get \((y+3/2)^2=x+41/4 - 5\), which simplifies to \((y+3/2)^2=x+21/4\).

Step 3: Identify Vertex and Direction

The equation, \((y+3/2)^2=x+21/4\), shows a parabola with vertex at \((-3/2, -21/4)\) opening to the right since x is positively related to \(y^2\).

Step 4: Determine Domain and Range

For a parabola opening to the right, the domain is \([h, +\infty)\) and the range is \((-\infty, +\infty)\). For this case, the domain is \([-21/4, +\infty)\) and the range is \((-\infty, +\infty)\).

Step 5: Determine if the relation is a Function

Since every x-value corresponds with multiple y-values, this relation is not a function.