Interpreting equations graphically (example 2)
About this video
Sal shows that solving the equation f(t) = h(t), where f(t) = e^(2t) - 2t^2 and h(t) = 4 - 5t^2, is the same as finding the t-values where the graphs of y = f(t) and y = h(t) intersect. Reading the graph, the solutions are t1 and t4, the t-coordinates of the intersection points.
You'll be able to
- explain why the solutions of f(t) = h(t) are the t-values where the graphs of f and h intersect.
- read the solutions of an equation from the intersection points of two graphs.
- tell that the solutions are the t-coordinates, not the y-coordinates, of the intersection points.
Key words
- Point of intersection
- A point where two graphs cross, so both functions give the same output there.
- Solution of an equation
- A value of the variable that makes both sides of the equation equal.
Transcript
- [Voiceover] Let f of t equal either the 2t minus 2t squared, and h of t equal four minus 5t squared. The graphs of y equals f of t, and y equals h of t, are shown below. So y equals f of t is here in green, so this is really y is equal to e to the 2t minus 2t squared.
We see f of t right over there. And y is equal to h of t is shown in yellow. All right. Now below that, they say, "Which of the following "appear to be solutions of e to the 2t minus 2t squared "equals four minus 5t squared? "Select all that apply." And I encourage you to pause the video and try to think about it.
Now the key here is to realize that either the 2t minus 2t squared, that was f of t. And four minus 5t squared is h of t. So another way of thinking about it, select all of the Ts for which f of t is equal to h of t. So all of the Ts where f of t is equal to h of t, well that's going to happen at the points of intersection.
So for example, at t1, we see at this point right here, t1 y1. So this tells us f of t1 is equal to h of t1, which is equal to y1. So f of t is going to be equal to h of t at t is equal to t1. We see that there because it's a point of intersection. Now let's keep on going.
Well they have another point of intersection right over here at t4, t4 y4. If you took f of t4, you're going to get y4. Or if you take h of t4, you're going to get y4. So f of t4 is equal to h of t4. f of t4 is equal to h of t4. So if you took e to the two times t4 minus two t4 squared, and all the way, that's going to equal to four minus five times t4 squared.
Since f of t and h of t equal each other, when t is equal to t4, these two things are going to equal each other, when t is equal to t4. And those are the only ones that are at a point of intersection. And so I think we are done. I can check my answer, and I got it right.
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