Problem 104 - Intermediate Algebra

Question

Under what conditions will the graph of \(x=a(y-k)^{2}+h\) have no \(y\) -intercepts?

Step-by-Step Solution

Answer

The graph has no y-intercepts when \(h > 0\).

Step 1: Understanding the Standard Parabolic Form

The given equation is in the form of a sideways parabola, written as \(x = a(y - k)^2 + h\). Here, \(h\) is the x-coordinate of the vertex, and \(k\) is the y-coordinate of the vertex. This form indicates that the parabola opens either to the right or left depending on the sign of \(a\).

Step 2: Determine when the Parabola has No Y-Intercepts

For a parabola to have no y-intercepts, it must never intersect the y-axis. This occurs when the entire parabola is positioned to the right or left of the y-axis. Since \(x = 0\) must be possible for a y-intercept, we set \(x = 0\) in the equation: \[x = a(y - k)^2 + h = 0\] and solve for \(y\). If this equation has no real solutions, then there are no y-intercepts.

Step 3: Analyzing the Equation for Real Solutions

Rearrange the equation to find the condition: \[0 = a(y - k)^2 + h\]Therefore, \[a(y - k)^2 = -h\]For real solutions for \(y\), \(a(y - k)^2\) must be non-negative because a square is always non-negative. Hence, if \(-h\) is negative (\(h > 0\)), the equation has no real solution.

Step 4: Conclusion on Conditions for No Y-Intercepts

The graph of the equation \(x = a(y-k)^2 + h\) will have no y-intercepts when the term \(h\) is positive, i.e., \(h > 0\). When \(h > 0\), the vertex of the parabola and the entire graph stay on the right side of the y-axis, thus avoiding any intersection.

Key Concepts

Parabolic Form

The equation presented, \(x = a(y-k)^{2} + h\), is in the parabolic form. This form describes a special kind of function known as a sideways parabola. Here's what makes it unique:

• Instead of opening upwards or downwards like a conventional parabola, this sideways version opens left or right.
• The direction in which it opens is determined by the sign of \(a\). If \(a\) is positive, the parabola opens to the right, and if it’s negative, it opens to the left.
• The terms \(h\) and \(k\) influence where the vertex (the parabola's peak or lowest point) is located. Specifically, \(h\) represents the x-coordinate, while \(k\) is the y-coordinate of the vertex.

Understanding the orientation and location of a parabola is crucial when analyzing its graph, especially when determining intercepts. This specific form helps students visualize and grasp the essence of why a parabola behaves the way it does graphically.

Y-Intercepts

A y-intercept is where a graph crosses the y-axis. For a parabola described in its sideways form \(x = a(y-k)^{2} + h\), finding this intercept means figuring out where \(x = 0\). Here's what that involves:

• You must set \(x = 0\) in the equation, which converts it into \(0 = a(y-k)^{2} + h\).
• From here, you rearrange it to: \(a(y-k)^{2} = -h\).

The crux is that the left-hand side, \(a(y-k)^{2}\), represents a square, and squares are always \(\geq 0\). Thus, if \(-h\) is negative (meaning \(h > 0\)), the equation won't equate to the necessary non-negative value, and hence, there will be no real solutions.
This technical mechanism signifies that the parabola doesn't intersect the y-axis. A formation of such indicates visually that the energy of the curve is restrained either completely to the left or to the right of the axis.

Real Solutions

When dealing with real solutions in our equation \(x = a(y-k)^{2} + h\), the goal is to assess when real y-values exist that satisfy the expression. To have real solutions for \(y\), the expression \(a(y-k)^{2}\) should be non-negative, aligning with the non-negativity of squares. Here's a deeper look:

• In our adjusted equation, \(a(y - k)^{2} = -h\), the condition for real solutions focuses on \(-h\) aligning with the non-negativity of the squared term.
• If \(-h\) is positive, it's compatible with the typical non-negative square outcome, suggesting possible y-values exist, i.e., solutions.
• Conversely, if \(h > 0\), then \(-h\) becomes negative, presenting a discord with non-negative outcomes of squares.

Thus, having no real solutions in this context indicates \(h\) being positive creates a scenario where the equation cannot be satisfied by any real numbers, establishing specific non-intersection scenarios on the graph.