Problem 54 - College Algebra

Question

Make a rough sketch in a rectangular coordinate system of the graphs representing the equations in each system. The system, whose graphs are a line with negative slope and a parabola whose equation has a negative leading coefficient, has one solution.

Step-by-Step Solution

Answer

The line with a negative slope could be represented by the equation \(y = -x + 3\) and the parabola with a negative leading coefficient could be represented by the equation \(y = -x^2 + 2x + 1\). The point where these two graphs intersect is the solution to the system of equations.

Step 1: Draw the first graph representing the line with a negative slope

Start by marking your axes. Next, because the line has a negative slope, it should decrease as it moves from left to right along the x-axis. An example equation might be \(y = -x + 3\), where we start the line at the y-intercept of 3 and drop one unit for each unit moved to the right on the x-axis because the slope is -1.

Step 2: Draw the second graph representing the parabola with a negative leading coefficient

Uniquely, a quadratic equation \(y = -x^2 + 2x + 1\) could be a representative of this category. The vertex form of a parabola's equation, \(y=a(x-h)^2+k\), could be used to find the vertex (h,k), axis of symmetry, direction of opening(up/down) based on 'a' being positive/negative. With a negative leading coefficient, the parabola will open downwards.

Step 3: Identify the point of Intersection

The point where the line and the parabola meet or intersect is the solution to the system of equations. As per the problem statement, these two graphs intersect at only one point, which means there is one solution. If graphed accurately, the point of intersection will be visible on the graph.