Problem 31 - Precalculus with Limits

Question

In Exercises 23-32, find the \( x \)- and \( y \)-intercepts of the graph of the equation. \( y^2 = 6-x \)

Step-by-Step Solution

Answer

The x-intercept of the graph of the equation is \(x = 6\). The y-intercepts are \(y = \sqrt{6}\) and \(y = -\sqrt{6}\).

Step 1: Determine the x-intercepts

To find the \(x\)-intercepts, we substitute \(y = 0\) into the equation and solve for \(x\). So, we get \( (0)^2 = 6 - x \), which simplifies to \(x = 6\).

Step 2: Determine the y-intercepts

To find the \(y\)-intercepts, we substitute \(x = 0\) into the equation and solve for \(y\). So, we get \(y^2 = 6 - 0\), which simplifies to \(y^2 = 6\). Therefore, \(y = \sqrt{6}\) and \(y = -\sqrt{6}\).

Key Concepts

Graph of an Equation

The graph of an equation represents all the points that satisfy the equation. To visualize this, imagine a coordinate plane where each point has an x-value (horizontal position) and a y-value (vertical position). For the equation \( y^2 = 6 - x \), the graph includes all points where, when you substitute the point's coordinates into the equation, you get a true statement.

When analyzing graphs, x-intercepts and y-intercepts are crucial. The x-intercepts are where the graph crosses the x-axis (horizontal). To find them, we set \( y = 0 \) because on the x-axis, the y-value of all points is zero. Similarly, to find the y-intercepts (where the graph crosses the y-axis), we set \( x = 0 \) because all points on the y-axis have an x-value of zero. This is particularly helpful for sketching the graph without plotting numerous points.

Solving Equations

The process of solving equations is about finding the value(s) that make the equation true. In our example, we're looking for values that, when substituted into \( y^2 = 6 - x \), yield a valid statement. To find the x-intercepts, we solved the equation with \( y = 0 \) to determine where the graph touches the x-axis, resulting in \( x = 6 \). Similarly, to find the y-intercepts, we set \( x = 0 \) and solved for \( y \) to find out where the graph meets the y-axis, yielding \( y \)= \( \pm\sqrt{6} \).

Solving equations can involve various operations such as adding, subtracting, multiplying, dividing, and even taking roots, as we did to find the y-intercepts. The key is to isolate the variable you are solving for by performing the same operations on both sides of the equation until you reach a solution.

Quadratic Equations

A quadratic equation is a type of polynomial equation of the second degree, generally taking the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \) , and \( c \) are constants, and \( a \) is not zero. The simple equation we're dealing with, \( y^2 = 6 - x \), can be seen as a quadratic equation when rearranged into standard form \( x + y^2 - 6 = 0 \).

Quadratic equations are unique because they can have up to two real solutions, as demonstrated by the y-intercepts in this example (\( \pm\sqrt{6} \)). The solutions to a quadratic equation are often found using methods such as factoring, the quadratic formula, or completing the square. In our problem, we found the intercepts directly, but for more complex quadratics, these methods become crucial tools for finding the points where a graph intersects the axes, as well as for understanding the shape and position of the parabola they form.