Problem 46 - Precalculus with Calculus Previews

Question

Find the \(x\) - and \(y\) -intercepts of the given parabola. \((x-1)^{2}=-2(y-1)\)

Step-by-Step Solution

Answer

The x-intercepts are \((1 + \sqrt{2}, 0)\) and \((1 - \sqrt{2}, 0)\). The y-intercept is \((0, \frac{1}{2})\).

Step 1: Understand Intercepts

The x-intercept occurs where the parabola crosses the x-axis, meaning y = 0. The y-intercept occurs where the parabola crosses the y-axis, meaning x = 0.

Step 2: Determine the x-intercepts

Set y = 0 in the equation \((x-1)^{2}=-2(y-1)\) to find the x-intercepts.Substitute y = 0:\((x-1)^{2} = -2(0-1)\) \((x-1)^{2} = 2\)Solve for x:\[x-1 = \pm \sqrt{2}\]\[x = 1 \pm \sqrt{2}\]Thus, the x-intercepts are \((1 + \sqrt{2}, 0)\) and \((1 - \sqrt{2}, 0)\).

Step 3: Determine the y-intercept

Set x = 0 in the equation \((x-1)^{2}=-2(y-1)\) to find the y-intercept.Substitute x = 0:\((0-1)^{2} = -2(y-1)\)\(1 = -2(y-1)\)Solve for y:\[y-1 = -\frac{1}{2}\]\[y = \frac{1}{2}\]Thus, the y-intercept is \((0, \frac{1}{2})\).

Key Concepts

x-intercept

The x-intercept of a parabola is the point or points where the graph crosses the x-axis. At this point, the value of y is always zero because the graph is not above or below the x-axis, it's right on it. Calculating the x-intercept involves solving the equation when y is set to zero.
To find the x-intercept, consider the equation of the parabola given:

• Firstly, replace y with 0 in the equation: \((x-1)^{2}=-2(0-1)\) which simplifies to \((x-1)^{2} = 2\).
• Next, solve the equation by taking the square root of both sides: \(x-1 = \pm \sqrt{2}\). This results in two potential solutions:
• Add 1 to both potential solutions to isolate x: \(x = 1 \pm \sqrt{2}\).

This calculation tells us there are two x-intercepts:

• \((1 + \sqrt{2}, 0)\)
• \((1 - \sqrt{2}, 0)\)

As a result, the parabola intersects the x-axis at these two points.

y-intercept

The y-intercept of a parabola is the point where the graph crosses the y-axis. Here, the x-coordinate is always zero. Finding the y-intercept involves setting x to zero in the parabola's equation and solving for y.
By substituting x = 0 in the equation \((x-1)^{2}=-2(y-1)\), we can find the y-intercept:

• Start substituting x with 0: \((0-1)^{2} = -2(y-1)\) which simplifies to \(1 = -2(y-1)\).
• Now, rearrange this to solve for y by first distributing the -2: \(y-1 = -\frac{1}{2}\).
• Finally, add 1 to both sides to isolate y: \(y = \frac{1}{2}\).

The y-intercept is thus at the point \((0, \frac{1}{2})\), meaning the parabola meets the y-axis at this vertical position.

quadratic equations

Quadratic equations are fundamental mathematical expressions, typically structured in the form \(ax^{2} + bx + c = 0\). These equations can form several shapes on a graph, one of the most common being a parabola.
Parabolas are distinct U-shaped curves which can open either upward or downward. The equation given, \((x-1)^{2}=-2(y-1)\), represents a parabola that has been rotated, signifying some transformation of the standard parabola. Key terms related to quadratic equations include:

• Vertex: The highest or lowest point on a parabola.
• x-intercepts: The points where the parabola crosses the x-axis, found by setting y to zero.
• y-intercept: The point where the parabola crosses the y-axis, determined by setting x to zero.
• Axis of symmetry: A vertical line through the vertex, splitting the parabola into two mirror images.

Understanding how to manipulate and graph these equations allows you to see their real-world applications, from projectile motions to reflecting light in telescopes. Reinterpreting these equations when they take a format such as \((x-1)^{2}=-2(y-1)\), reveals the distance from a certain reference point (like the origin) and helps locate where intercepts occur on the graph.