Problem 34 - Algebra 2

Question

Graph each equation. $$ y^{2}-x-2 y+1=0 $$

Step-by-Step Solution

Answer

The graph of the given equation is a parabola with the vertex at (2, 1) and it opens to the left.

Step 1: Rewrite the Equation

First, group the y-terms together and add x to both sides of the equation. Doing so, the equation will be rewritten as \(y^{2}-2y=-x+1\).

Step 2: Complete the Square

To complete the square for \(y^2 - 2y\), take half the coefficient of y, square it and add it to both the sides of the equation. Half of -2 is -1, and \(-1^2 = 1\). Therefore, the completed square will be \(y^2 - 2y + 1 = -x + 2\). This can be written as \((y-1)^2 = -x+2\).

Step 3: Rewrite in Standard Form

Now rewrite the equation as \(x = -(y-1)^2 + 2\). This equation is in the form of a parabola \(x = - 4a(y-h)^2 + k\), where a = 1/4, h = 1 and k = 2.

Step 4: Identify Key Components

The vertex of the parabola is at the point (k, h) = (2, 1). The parabola opens to the left because the coefficient of \((y-1)^2\) is negative.

Step 5: Draw the Graph

The last step is to draw the graph. Start by marking the vertex. Since the parabola opens to the left, the points on the parabola would be formed by decreasing the x-value while keeping the y-values the same.

Key Concepts

Completing the Square

Completing the square is a helpful algebraic technique used to rewrite quadratic expressions in a form that makes them easier to solve or graph. When you have a quadratic in the form of \(y^2 - 2y\), the idea is to transform this into a perfect square trinomial.

Here's how it works:

• Take the coefficient of \(y\), which is -2 in this case.
• Divide it by 2, which gives you -1.
• Square this result to get 1.
• Add and subtract this square (1) inside the equation for balance.

So \(y^2 - 2y\) becomes \((y - 1)^2\) after including the added 1.

This technique transforms complex equations into a format that is simpler to manipulate, enabling us to identify characteristics like the vertex of a parabola more easily.

Parabola Vertex

The vertex of a parabola is a crucial point that indicates its peak or lowest point, depending on the parabola's direction. For an equation like \((y-1)^2 = -x + 2\), the vertex forms the center of the parabola's symmetry. In our rewritten equation \(x = -(y-1)^2 + 2\), the vertex is at \((k, h) = (2, 1)\).

Here's how these values come into play:

• \(h\) is the value subtracted from \(y\) inside the square, which is 1.
• \(k\) is the value added to or subtracted from the other side of the equation, which is 2.

The coordinates (2, 1) provide the exact point on the graph where the vertex resides. Understanding this point helps anchor the rest of your graphing efforts, as all lines around the vertex can be plotted in relation to it.

Parabola Direction

The direction in which a parabola opens is directly related to the coefficient in front of the squared term. For our example, \(x = -(y-1)^2 + 2\), the coefficient of \((y-1)^2\) is -1.

This tells us a couple of key things:

• If the coefficient is negative, as it is here, the parabola opens to the left (or downward if \(x\) and \(y\) roles are reversed).
• If it were positive, the parabola would open to the right.

Knowing this direction is vital for sketching a realistic graph and understanding how the quadratic function behaves. It affects how you plot additional points relative to the vertex and can determine the nature of any solutions or intersections with the axes.

Graphing Equations

Graphing quadratic equations like parabolas involves plotting points and drawing symmetric curves. With the equation \(x = -(y-1)^2 + 2\), start at the vertex, which is the point \((2, 1)\).

Follow these steps for an accurate graph:

• Plot the vertex point on your graph.
• Since the parabola opens to the left, choose values of \(y\) close to the vertex and calculate corresponding \(x\) values to find additional points.
• Remember the symmetry around the vertex as you plot these points.
• Draw the curve smoothly connecting these points, ensuring it opens leftward as indicated by the equation.

Plotting a well-drawn parabola helps visualize solutions, intersections, and other features of the quadratic function, providing a deeper understanding of its properties.