Problem 47 - Algebra 2

Question

Graph each equation. $$ y^{2}-8 y+8 x=-16 $$

Step-by-Step Solution

Answer

The graph of the equation \( y^{2}-8 y+8 x=-16 \) is a rightward opening parabola having its vertex at the point (0,4).

Step 1: Convert Equation to Suitable form

The given equation is \( y^{2}-8 y+8 x=-16 \). The first step is to rearrange it in the form of \( y \). To do this, 16 is added to both sides of the equation. Consequently, the equation becomes \( y^{2}-8y+16 = 8x \).

Step 2: Recognize the Form

After rearrangement, it becomes clear that the equation \( y^{2}-8y+16 = 8x \) is the standard form of a parabola equation \( (y-h)^2=4ax \), where \( (h,k) \) are the coordinates of the vertex of the parabola and \( a \) is a constant. This equation is also equivalent to \( y^2 = 2 \cdot 4 \cdot x \). Thus, \( h=4 \), \( k=0 \), and \( a=2 \).

Step 3: Graph Equations

The parabola opens to the right if \( a \) is positive and to the left if \( a \) is negative. Here, \( a \) is positive. Hence, the parabola opens towards the right. Correspondingly, plot the vertex at \( (0,4) \). Then, because \( a=2 \), move 2 units to the right and 1 unit upward from the vertex to plot a point, and 1 unit downward to plot another point. Repeat this process to plot additional points. Finally, sketch a smooth curve passing through those points.

Key Concepts

Standard Form of Parabola

The standard form of a parabola is a way to express the equation of a parabola so that key features are easy to identify. For a parabola that opens horizontally, the standard form is \((y - k)^2 = 4a(x - h)\). Here are the critical components:

• \((h, k)\) – the coordinates of the vertex; the point where the parabola is most "centered".
• \(a\) – a constant that tells you how "wide" or "narrow" the parabola is.

In our problem, we've rearranged the equation to \(y^2 - 8y + 16 = 8x\), which matches the form \((y - 4)^2 = 8x\). This reveals the vertex and how the parabola behaves.

Vertex of Parabola

The vertex of a parabola is a crucial point that represents its maximum or minimum, depending on the orientation. It serves as the "tip" of the curve. From the standard form \((y - k)^2 = 4a(x - h)\), the coordinates of the vertex can be directly seen as \((h, k)\).

In this exercise, after rearranging the equation, we identify the vertex as \((0, 4)\). This point provides a reference for graphing, indicating where the parabola changes direction.

Graphing Parabolas

Graphing a parabola involves using its vertex and direction. Starting with the vertex gives you a key anchoring point. Here’s how you can proceed:

• Plot the vertex \((0, 4)\) on the graph.
• Use the value of \(a\) to determine how to plot additional points. A positive \(a\) means moving to the right for a horizontally opening parabola.
• From the vertex, go "\(a\) units" horizontally (right or left), and then up or down, to place more points.

By connecting these plotted points smoothly, you’ll get the shape of the parabola.

Opening of Parabola

The direction in which a parabola opens is determined by the sign of \(a\) in its standard form. For horizontally opening parabolas, like in our scenario:

• If \(a > 0\), the parabola opens to the right.
• If \(a < 0\), it opens to the left.

In our problem, since \(a = 2\), the parabola opens to the right, meaning it curves away from the vertex in the positive x-direction. This characteristic is vital when sketching the parabola's shape.