Problem 74 - Algebra for College Students

Question

Graph each of the following equations by solving for \(y\) a) \(x=y^{2}-1\) b) \(x=-y^{2}\) c) \(x^{2}+y^{2}=4\)

Step-by-Step Solution

Answer

Solve each for y, resulting in: a) \(y = \pm\sqrt{x+1}\), b) \(y = \pm\sqrt{-x}\), c) \(y = \pm\sqrt{4 - x^2}\). Graph accordingly.

Step 1 (Equation a): Solve for y

Given the equation: \(x = y^2 - 1\). To solve for \(y\), isolate \(y^2\): \(y^2 = x + 1\). Take the square root of both sides: \(y = \pm\sqrt{x + 1}\). This results in two functions: \(y = \sqrt{x + 1}\) and \(y = -\sqrt{x + 1}\).

Step 2 (Equation a): Graph the functions

Plot the graphs of \(y = \sqrt{x + 1}\) and \(y = -\sqrt{x + 1}\). These are upward and downward parabolas respectively, opening to the right.

Step 3 (Equation b): Solve for y

Given the equation: \(x = -y^2\). Isolate \(y^2\) by multiplying both sides by -1: \(y^2 = -x\). Then take the square root of both sides: \(y = \pm\sqrt{-x}\). This results in two functions: \(y = \sqrt{-x}\) and \(y = -\sqrt{-x}\).

Step 4 (Equation b): Graph the functions

Plot the graphs of \(y = \sqrt{-x}\) and \(y = -\sqrt{-x}\). These are upward and downward parabolas respectively, opening to the left.

Step 5 (Equation c): Solve for y

Given the equation \(x^2 + y^2 = 4\). Solve for \(y^2\) by isolating it: \(y^2 = 4 - x^2\). Take the square root of both sides: \(y = \pm\sqrt{4 - x^2}\). This results in two functions: \(y = \sqrt{4 - x^2}\) and \(y = -\sqrt{4 - x^2}\).

Step 6 (Equation c): Graph the functions

Plot the graphs of \(y = \sqrt{4 - x^2}\) and \(y = -\sqrt{4 - x^2}\). These are semicircles with radius 2, the first opening upwards, and the second downwards, forming a circle centered at the origin.

Key Concepts

solving for y

When dealing with equations, sometimes we aim to solve for a specific variable, such as y. This involves rearranging the equation to isolate y on one side.

For example, given the equation: \(x = y^2 - 1\), we want to isolate y. This means we need to get y by itself on one side of the equation.
Here's how to do it step-by-step:

• Add 1 to both sides: \(x + 1 = y^2\)
• Then take the square root of both sides: \(y = \pm \sqrt{x + 1}\)

This gives us two solutions: \(y = \sqrt{x + 1}\) and \(y = -\sqrt{x + 1}\). Don't forget, taking the square root provides both positive and negative roots due to the nature of quadratic equations.
These resulting functions can be graphed to visualize the solutions.

parabolas

A parabola is a U-shaped curve that can open up, down, left, or right. Parabolas are graph representations of quadratic equations and can have various orientations.

For instance, in the equation \(x = y^2 - 1\), we rearranged to get \(y = \pm \sqrt{x + 1}\). These functions, \ y = \sqrt{x + 1}\ and \ y = -\sqrt{x + 1}\, are upward and downward parabolas that open to the right.

Similarly, the equation \(x = -y^2\) gives us \(y = \pm \sqrt{-x}\) after rearranging. These functions, \ y = \sqrt{-x}\ and \ y = -\sqrt{-x}\, form parabolas that open to the left.

Remember that the direction in which a parabola opens depends on the sign and position of the squared term.

semicircles

A semicircle is half of a circle. When graphing equations involving \(x^2 + y^2 = r^2\), they can form full circles if both positive and negative square root solutions are considered.

Let's look at \(x^2 + y^2 = 4\). Solving for y:

• First, rearrange: \(y^2 = 4 - x^2\)
• Then, take the square root: \(y = \pm \sqrt{4 - x^2}\)

This gives us two functions: \ y = \sqrt{4 - x^2}\ and \ y = -\sqrt{4 - x^2}\.

Graphing these reveals two semicircles. One represents the top half (upper semicircle) and the other represents the bottom half (lower semicircle) of a circle with radius 2, centered at the origin.

square roots

Square roots are used when solving any equation where a variable is squared. It's essential to remember that the square root of a number has two values: one positive and one negative.

For example, in the equation \(x = y^2 - 1\), if we solve for y, we get: \(y = \pm \sqrt{x + 1}\). This reveals both the positive and negative roots, meaning for every x, there are two corresponding y-values.

Likewise, with the equation \(x = -y^2\), solving for y yields: \ y = \pm \sqrt{-x}\. Again, we see both positive and negative roots.

Understanding square roots is crucial for solving and graphing quadratic equations, as it ensures capturing all possible solutions.