2.07 Quadratic systems

Graph the equations on the same coordinate plane.

The solution(s) to a system of equations can be represented graphically as their point(s) of intersection. We can use technology to graph the two equations, or, if drawing them by hand, it will be useful to first fill out a table of values for both equations. We can use what we know about function types to pick the best range of table values. For example, we know the vertex of the quadratic equation will be at x = \dfrac{2}{\left(2\right)\left(1\right)} so we will want to choose x-cordinates on either side of x=1.

Notice that from the table of values both functions have the points \left(-2,5\right) and \left(3,0\right). We want these points of intersection to be visible on our graph. We also want the vertex of the parabola, \left(1,-4\right), to be visible.

We want to ensure that the x-values on our graph cover at least the interval -3 \leq x \leq 4 and the y-values on our graph cover at least the interval -5 \leq y \leq 6.

Identify the coordinates of the solution(s) to the system of equations.

The points of intersection occur at \left(-2, 5\right) and \left(3, 0\right). When creating the table, we saw that these values made both equations true, so these coordinate pairs are the solutions to the system of equations.

Alternatively, we could have solved the system of equations algebraically, by equating both equations and solving for x.

Using the zero product property, we can see the two solutions to this new quadratic are x=-2 and x=3.

We can now substitute these into one of the given equations to find the corresponding y-values.

This corresponds with the solution \left(-2,5\right).

This corresponds with the solution \left(3,0\right).

Determine the number of solutions to the system of equations.

To determine the number of solutions to the system of equations, we can graph the system on a coordinate plane and look for points of intersection.

Find the solution(s) to the system of equations.

From the graph, we see that one of the solutions is \left(1,2\right) but the other solution does not have integer coordinates. For this reason, we must use technology to estimate the coordinates of the other point or solve algebraically.

When solving algebraically, using the substitution method is usually easiest, but the elimination method is also valid.

Let's begin by numbering the equations to make them easier to work with.

Since both equations already have y isolated, we can start by substituting equation 1 into equation 2 to eliminate y from the equation.

Now, we can solve for x using the zero product property.

Finally, we will substitute the x-values back into one of the equations to solve for y.

The solutions to the system are \left(\dfrac{1}{3},\dfrac{34}{9}\right) and \left(1,2\right).

We can verify these solutions with technology or by substituting the values back into the equations to see if the solutions make both equations true.

Since we solved the system algebraically, we will check the answer by graphing with technology and using the tracing tool.

Write a system of equations to model this problem.

Notice that the base jumper's trajectory follows a linear path in the form y = mx+b.

The ball's trajectory follows a quadratic path in the form y=a(x-h)^2+k.

For the base jumper, their initial height is 560 ft. and they are falling at a rate of -25 \text{ ft/s}. Using a linear model, the height, y, after x seconds, is given by: y=-25x+560.

We know the ball follows a path in the form a(x-h)^2+k where a is the force of gravity which is -16 \text{ ft/s}^2. Since it reaches its maximum of 1250 ft. after 1.5 seconds, we know h=1.5 and k=1250. Therefore, the ball's height can be modeled by the function: y=-16\left(x-1.5\right)^{2}+1250.

Graph the height of the base jumper and ball on the same coordinate plane.

We can use technology to obtain a graph of the two functions, using the context and graphs to identify an appropriate domain and range for the graph.

Alternatively, we can graph by hand using key features of the graph such as:

• The equation for the base jumper's height is a linear function, so we can identify the y-intercept, find another point on the line using the slope, and then graph the line through those two points.

• The equation for the ball is in vertex form, so we can see the direction of opening from the coefficient, plot the vertex, then plot the y-intercept of \left(0,1214\right), and substitute another value for x to get the shape.

Consider what domain is appropriate for each graph in the context. The graphs should not extend beyond their valid domain.

Determine the time the base jumper and ball are at the same height.

Let's use technology to solve this problem. We can start by inputting the two equations for the base jumper and ball into a graphing calculator. We already have a sketch, so know we should end up with a parabola and a line.

We can estimate the coordinates by eye, but we should determine the coordinates more precisely using an intersection tool.

From the graph, we can see that the graphs intersect at a single point in the domain appropriate to the context. To find the coordinates of this point, we can use the intersection tool. For the built-in GeoGebra graphing calculator, we need to click on the point of intersection.

The point where the graphs intersect within the restricted domain of the context is:\left(9.07, 333.26\right).

So, the ball and base jumper will be at the same height of approximately 333.26 ft at 9.07 seconds after the throw/jump.

We can confirm this algebraically as well:

The solutions of this quadratic equation would give us the x-coordinates for any points of intersection between the two graphs.

The discriminant of this equation is \left(73\right)^{2}-4 \cdot \left(-16\right) \cdot 654=47\,185 \gt 0 which means there are two real solutions.

We could then use the quadratic formula to get the x-values of the points of intersection, ignoring negative solutions for this context.