Question

Teresa graphs the following 3 equations: y=2x , y=x²-2 , and y=2x² . She says that the graph of y=2x will eventually surpass both of the other graphs. Is Teresa correct? Why or why not?

1) Teresa is correct. The graph of y=2x grows at an increasingly increasing rate, but the graphs of y=x²-2 and y=2x² both grow at a constantly increasing rate. Therefore, the graph of y=2x will eventually surpass both of the other graphs.
2) Teresa is not correct. The graph of y=2x grows at an increasing rate and will eventually surpass the graph of y=x²-2. However, it will never surpass the graph of y=2x² because the y-value is always twice the value of x².
3) Teresa is not correct. The graph of y=2x² already intersected and surpassed the graph of y=2x at x=1. Once a graph has surpassed another graph, the other graph will never be higher.
4) Cannot be determined based on the given information.

Answer 1

Teresa's claim that y=2x will surpass both y=x²-2 and y=2x² is incorrect because while y=2x will surpass y=x²-2 at large values of x, it will never surpass y=2x² which grows at a rate proportional to x squared.

Step-by-step explanation:

Teresa is not correct in her assertion that the graph of y=2x will eventually surpass both of the other graphs. The equation y=2x represents a linear function with a constant growth rate, meaning that for each unit increase in x, y increases by 2 units. On the other hand, the equations y=x²-2 and y=2x² represent quadratic functions, which exhibit an increasing growth rate as x becomes larger. Specifically, the graph of y=2x will surpass y=x²-2 for sufficiently large values of x because the quadratic term becomes negligible compared to the linear growth. However, since y=2x² has a growth rate that is proportional to the square of x, the value of y for this function will always be greater than for the linear function y=2x for all x greater than 0.