Problem 70 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine any point(s) of intersection algebraically. Then verify your result numerically by creating a table of values for each function. $$\begin{aligned} &y=-x^{2}+3 x+1\\\ &y=-x^{2}-2 x-4 \end{aligned}$$

Step-by-Step Solution

Answer

The point of intersection of the two given functions is (-1, -3). This means, for both functions, when we substitute x = -1, we get y = -3.

Step 1: Set Both Functions Equal

To find the intersection point of two functions, we should first set them equal to each other. So, we will have: \[-x^{2}+3 x+1 = -x^{2}-2 x-4\] As -x^2 cancels out from both sides of the equation.

Step 2: Solve for x

Rearrange the equation to solve for x by moving terms related to x to one side and constant to the other. This will leave us with: \(5x + 5 = 0\). Dividing through by 5, we find that \(x = -1\).

Step 3: Solve for y

Substitute the value of x into any of the original equations to get the y-coordinate of the intersection point. \(y = -(-1)^2 + 3*(-1) + 1 = -3\). Therefore, the point of intersection is (-1, -3).

Step 4: Verify Numerically

To verify the solution, create a table of values for each function around x = -1 and you will find that both y-values are equal to -3 at x = -1, thus confirming the solution.

Key Concepts

Intersection Point of Functions

When two functions are plotted on a coordinate plane, the point(s) where their graphs intersect are known as the intersection points. To find these points algebraically, we equalize the functions since at the intersection they share both x and y coordinates.

To illustrate, let's consider the functions \(y = -x^2 + 3x + 1\) and \(y = -x^2 - 2x - 4\). As the y values are the same at the intersection, we can set the right-hand sides of the equations equal to each other and solve for x, leading us to the intersection point.

By simplifying the equation, we remove identical terms present on both sides. This way, we're left with an easier equation representing the difference between two functions. For our example, when the term \( -x^2 \) cancels out, we effectively compare the remaining linear parts that are unique to each function, leading us to the solution for x.

Quadratic Functions

Quadratic functions take the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and the graph of a quadratic function is a parabola. When a is positive, the parabola opens upwards, and when a is negative, as with our functions, it opens downwards.

One of the key features of these parabolas is their symmetry around the vertical line called the axis of symmetry, which passes through the vertex, the highest or lowest point of the parabola. The symmetry can be particularly helpful when we're trying to verify intersection points because symmetrical points will have the same y-values for a given x-value either side of the vertex.

This characteristic makes quadratics predictable and easier to analyze, especially when we're looking to understand the behavior of the graph near the intersection points.

Algebraic Verification

Algebraic verification is a pivotal step that ensures we have correctly solved equations, particularly when we find intersection points or any solution to algebraic problems. It involves substituting the found solution back into the original equations to confirm its validity.

For our functions, after determining \(x = -1\), we substitute it back into one of the original equations to find the corresponding y-value. By obtaining \(y = -3\), we verify that the point of intersection is indeed \( (-1, -3) \).

Going beyond just algebraic substitution, we also create a table of values for each function around the suspect x-value. By scrutinizing the behavior of both functions around x = -1, we can numerically see that they both yield the same y-value at this point, further confirming the accuracy of our algebraic solution.