Problem 29 - Beginning and Intermediate Algebra

Question

For each equation, identify the vertex, axis of symmetry, and \(x\) - and \(y\) -intercepts. Then, graph the equation. $$x=-(y-4)^{2}+5$$

Step-by-Step Solution

Answer

Vertex: (5, 4) Axis of Symmetry: \(y = 4\) x-intercept: (-11, 0) y-intercepts: \((0, 4 + \sqrt{5})\) and \((0, 4 - \sqrt{5})\) Graph the equation by plotting the vertex, axis of symmetry, and the intercepts, and sketching the parabola.

Step 1: Identifying the Vertex

Recall that the vertex form of a parabolic equation is: \(x = a(y - k)^2 + h\), where \((h, k)\) is the vertex of the parabola. The given equation is: \(x = -(y-4)^2 + 5\) So, comparing it to the standard vertex form, you can see that \(a = -1\), \(h = 5\), and \(k = 4\). Thus, the vertex of the parabola is \((5, 4)\).

Step 2: Identifying the Axis of Symmetry

The axis of symmetry is a horizontal line, since the parabola opens horizontally, and it goes through the vertex. In this case, the axis of symmetry is a horizontal line going through \(4\) on the y-axis, which can be written as the equation: \(y = 4\).

Step 3: Finding the x-intercepts

To find the x-intercepts, set \(y = 0\) and solve for \(x\): \(x = - (0 - 4)^2 + 5\) \(x = -(4)^2 + 5\) \(x = -16 + 5\) \(x = -11\) Thus, the only x-intercept for this equation is \((-11, 0)\).

Step 4: Finding the y-intercepts

To find the y-intercepts, set \(x = 0\) and solve for \(y\): \(0 = - (y - 4)^2 + 5\) \((y-4)^2 = 5\) \(y-4 = \pm \sqrt{5}\) \(y = 4 \pm \sqrt{5}\) Thus, the y-intercepts for this equation are \((0, 4 + \sqrt{5})\) and \((0, 4 - \sqrt{5})\).

Step 5: Graph the Equation

Now, we can graph the equation using the information we gathered in Steps 1-4: 1. Plot the vertex at \((5, 4)\). 2. Draw the axis of symmetry at \(y = 4\). 3. Plot the x-intercept at \((-11, 0)\). 4. Plot the y-intercepts at \((0, 4 + \sqrt{5})\) and \((0, 4 - \sqrt{5})\). 5. Sketch the parabola using these points and the axis of symmetry to ensure symmetry. Now, you should have a clear graph of the parabolic equation, \(x = -(y-4)^2 + 5\).

Key Concepts

Vertex

The vertex of a parabola is a crucial point as it represents the peak or the lowest point on the curve. For a parabola in vertex form, depending on its orientation, the vertex is defined as \((h, k)\). This form shows how the parabola shifts from the origin, moving horizontally by \(h\) and vertically by \(k\).
In our equation \(x = -(y-4)^2 + 5\), the vertex can be found by comparing it with the general vertex form \(x = a(y - k)^2 + h\). Here, \(a = -1\), \(h = 5\) and \(k = 4\). Hence, the vertex is located at \((5, 4)\).
The vertex not only marks a special point of the parabola but also helps in understanding the parabola's symmetry and how it opens. For horizontal parabolas like this one, the vertex indicates the most extreme left or right point.

Axis of Symmetry

The axis of symmetry is a line that runs through the vertex and divides the parabola into two equal, mirror-image parts.
In our equation, since it is a horizontally opening parabola, the axis of symmetry is a horizontal line.
To determine this line, we look at the vertex which is positioned at \((5, 4)\). The axis of symmetry aligns with the y-coordinate of the vertex, here \(y = 4\).
This horizontal line at \(y = 4\) ensures that if you were to fold the parabola along this line, both halves would perfectly overlap, showing the parabola's symmetry. Understanding this concept is crucial when sketching or graphing parabolas as it gives us insight into their balance and orientation.

x-intercepts

The x-intercepts of a parabola are points where the graph crosses the x-axis. At these intercepts, the value of \(y\) is zero.
Finding the x-intercepts involves setting \(y = 0\) in the equation and solving for \(x\).
In our specific case, substitute \(y = 0\) into \(x = -(y-4)^2 + 5\):

• \(x = -(0-4)^2 + 5\)
• \(x = -16 + 5\)
• \(x = -11\)

Thus, the only x-intercept is \((-11, 0)\).
This means that the parabola crosses the x-axis at this point, giving us a vital reference for sketching the graph and understanding its extent along the x-axis.

y-intercepts

The y-intercepts occur where the graph meets the y-axis, which happens when \(x = 0\).
To find them, solve the equation for \(y\) by setting \(x = 0\).
In our equation, substitute 0 for \(x\):

• \(0 = -(y-4)^2 + 5\)
• \((y-4)^2 = 5\)
• \(y - 4 = \pm \sqrt{5}\)
• \(y = 4 + \sqrt{5}\) and \(y = 4 - \sqrt{5}\)

Therefore, the y-intercepts are located at \((0, 4 + \sqrt{5})\) and \((0, 4 - \sqrt{5})\).
The presence of two y-intercepts implies that the parabola has its turning point in-between these intercepts, especially helpful when drawing the parabola, as they provide additional points of reference on the graph.