Problem 58 - College Algebra and Calculus: An Applied Approach

Question

Use a graphing utility to find the point(s) of intersection of the graphs. Then confirm your solution algebraically. $$\left\\{\begin{array}{l}y=-2 x^{2}+x-1 \\ y=x^{2}-2 x-1\end{array}\right.$$

Step-by-Step Solution

Answer

The two points of intersection of the given equations are (0, -1) and (1, -2).

Step 1: Graphical Solution

First, plot both of the provided equations on the same graph using graphing software. The points where the curves of both equations overlap are the solutions to this particular system of equations.

Step 2: Algebraic Solution - Set Equations Equal

To confirm the solution algebraically, one must set the two given equations equal to each other and solve for the variable x. \(-2x^2 + x - 1 = x^2 - 2x - 1\).

Step 3: Algebraic Solution - Simplify equation

Next, combine like terms to simplify the equation: \(-2x^2 - x^2 + x + 2x - 1 + 1 = 0 \) which simplifies to \(-3x^2 + 3x = 0\).

Step 4: Further Simplification and Solve for x

Factor the equation to its simplest form \(x(-3x + 3) = 0\). Solving for x gives \(x = 0\) or \(x = 1\).

Step 5: Solve for y

Finally, substitute the x-values back into either original equation to solve for y. For \(x = 0, y = -1\), and for \(x = 1, y = -2\). So the two points of intersection are (0, -1) and (1, -2).

Key Concepts

Graphical Solution

When solving systems of equations, one of the first techniques we can use is the graphical approach. This involves plotting the given equations on the same set of axes to visually identify their points of intersection.
In the problem at hand, we have two equations in quadratic form:

• \( y = -2x^2 + x - 1 \)
• \( y = x^2 - 2x - 1 \)

By using a graphing utility or tool, such as a graphing calculator or software like Desmos, these curves are plotted on a coordinate plane. The points where these two graphs cross each other are the solutions to the system of equations.
In this particular instance, the graphs intersect at two distinct points, giving us the values \((0, -1)\) and \((1, -2)\) where both equations are satisfied.

Algebraic Solution

After identifying the intersection points graphically, confirmation is often required through an algebraic process. This involves manipulating the equations to find solutions mathematically.
With the given system of equations:

• \( y = -2x^2 + x - 1 \)
• \( y = x^2 - 2x - 1 \)

To find the algebraic solution, we set the two equations equal to each other since both express \(y\):\(-2x^2 + x - 1 = x^2 - 2x - 1.\)
By combining like terms, simplifying the expression becomes much more manageable. We gather all terms to one side of the equation:\(-2x^2 - x^2 + x + 2x - 1 + 1 = 0.\)
This simplifies to:\(-3x^2 + 3x = 0.\)
By factoring out the greatest common factor, the equation becomes \(x(-3x + 3) = 0\). Solving this results in

• \(x = 0\)
• \(x = 1\)

Thus, providing the x-coordinates of the intersections.

Quadratic Equations

Quadratic equations are a fundamental part of algebra and mathematics in general. They are polynomial equations of degree two, typically expressed in the standard form \(ax^2 + bx + c = 0\). In our problem, both equations are quadratics:

• \(y = -2x^2 + x - 1\)
• \(y = x^2 - 2x - 1\)

These quadratics are parabolas (U-shaped curves) and can open upwards or downwards, depending on the sign of the \(x^2\) coefficient. If it's positive, the parabola opens upwards; if negative, downwards.
Solving quadratic equations can be done in multiple ways, such as:

• Graphing, as discussed, to find intersection points visually.
• Factoring, when the equation can be rewritten as a product of linear expressions.
• Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which works for any quadratic equation.

These methods allow us to solve for \(x\), and subsequently \(y\), to understand where and how these functions behave similarly or differently in their domain.