Problem 64 - Precalculus with Limits

Question

In Exercises 57-70, find any points of intersection of the graphs algebraically and then verify using a graphing utility. \(16x^2-y^2+16y-128=0\) \(y^2-48x-16y-32=0\)

Step-by-Step Solution

Answer

The algebraic solution gives the intersection points of the two curves. Then a graphing utility could be used to confirm these intersection points visually. Always remember to substitute the values of one variable (either \(x\) or \(y\)) back into one of the original equations to find the corresponding values of the other variable.

Step 1: Rearrange the Equations

First, let's rearrange the two given equations to isolate \(y^2\) in both equations such that they could be equated. These rearranged equations are: (1) \(y^2 = 16x^2+16y-128\) (2) \(y^2 = 48x+16y+32\)

Step 2: Set the Two Equations Equal and Solve

Next, set the two equations equal to each other and simplify it into a quadratic equation: \(16x^2 + 16y - 128 = 48x + 16y + 32\)Simplify it to obtain: \(16x^2 - 48x - 160 = 0\)This is a quadratic equation in variable \(x\). Its roots can be obtained using the quadratic formula \(x = {[-b ± sqrt(b^2 - 4ac)] / 2a }\) where a = 16, b = -48, and c = -160 in our equation.

Step 3: Finding the Value of y

After finding the values of \(x\), substitute them into one of the original equations to find the corresponding \(y\) values. For example, if we use the second equation \(y^2 - 48x - 16y - 32 = 0\), we can solve for \(y\) by rearranging the equation to \(y^2 + 16y + 48x - 32 = 0\) which is in the form of a quadratic equation. Again, use the quadratic formula to solve for \(y\).

Step 4: Using Graphing Utility

Finally, all the points of intersection of the two curves found in the previous steps should be verified visually using a graphing utility. Graph the two equations separately and see if the graph intersects at the points anticipated.

Key Concepts

Quadratic Equations

A quadratic equation is a polynomial equation of the second degree, typically in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. The graph of a quadratic equation is a parabola, which could open upwards or downwards depending on the sign of \(a\).

For example, in our exercise, we encountered the quadratic equation \(16x^2 - 48x - 160 = 0\) when we set the two rearranged equations equal to each other. To solve this, we can use various methods such as factoring, completing the square, or utilizing the quadratic formula. It is crucial to understand that a quadratic equation can have two real solutions, one real solution, or two complex solutions, depending on the discriminant value \(b^2 - 4ac\).

Graphing Utility Verification

Once the algebraic solution to a system of equations is found, verifying the solution with a graphing utility is a good practice to ensure accuracy. A graphing utility, such as a graphing calculator or computer software, visually represents equations on a coordinate plane.

By inputting our original equations and observing where the graphs intersect, students can visually confirm the points of intersection that they calculated algebraically. For the given exercise, intersecting parabolas from the equations \(16x^2-y^2+16y-128=0\) and \(y^2-48x-16y-32=0\) are graphed, and the points of intersection from the algebraic solution should match the points where the two parabolas cross on the graph. This visual verification acts as a reinforcement of the algebraic results.

Quadratic Formula

The quadratic formula is a fundamental tool in algebra for finding the roots of a quadratic equation. The formula is given by \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\), where \(a\), \(b\), and \(c\) are the coefficients of the equation \(ax^2 + bx + c = 0\). The term under the square root, \(b^2 - 4ac\), is known as the discriminant and it determines the nature of the roots.

In the provided exercise, we use the quadratic formula to solve for \(x\) in the equation \(16x^2 - 48x - 160 = 0\), where \(a = 16\), \(b = -48\), and \(c = -160\). We also apply it to find the corresponding \(y\) values after determining \(x\). Understanding how to apply the quadratic formula is crucial for solving these types of equations efficiently.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to solve for unknown variables. Key skills include expanding, factoring, combining like terms, and moving terms across the equals sign while maintaining the equation's balance.

In the solution steps provided, algebraic manipulation is used extensively. Initially, we rearrange the given equations to isolate the \(y^2\) term, allowing us to set the equations equal to each other. Subsequent simplification leads us to a solvable quadratic equation in \(x\). Further manipulation is required to solve for \(y\) after we obtain the \(x\) values. Mastery of algebraic manipulation techniques is vital for solving systems of equations and is a foundational skill for all algebra students.