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### Course: Algebra (all content) > Unit 5

Lesson 8: Solving equations by graphing (Algebra 2 level)- Interpreting equations graphically
- Interpreting equations graphically (example 2)
- Interpret equations graphically
- Solving equations graphically (1 of 2)
- Solving equations graphically (2 of 2)
- Solving equations graphically
- Solve equations graphically

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# Interpreting equations graphically (example 2)

Sal finds the coordinates of points on a graph that are solutions to a given equation.

## Want to join the conversation?

- Why couldn't the answer be y1?(29 votes)
- As Sal states at0:47the input variable is , not y. The graph only shows the corresponding y to the input t, not to an input y. So it's
`e^(2 * t1) - 2 (t1)^2 = y1 = 4 - 5 (t1)^2`

, with the result y1 but the solution t1.(17 votes)

- I have a question. The question asks the
**solution**to the equation. And every y value is a solution to each x input. So shouldn't y1 and y4 be the solution to the equation?(6 votes)- Read the statement carefully... You need solutions to the equation: e^(2t) - 2t^2 = 4 - 5t^2.

The only variable being used in this version of the equation is the variable "t". Thus, the solution(s) if you solve would be value(s) of t, not y. Select the values of "t" that make this equation true (both sides the same) and you get the values: t1 and t4.

The y1 and y4 represent the values that each side should become when you plug in the appropriate t-value. They aren't solutions to this version of the equation. They would be solutions to the original functions.

Hope this helps.(17 votes)

- I'm troubled by Marioland's issue as well. The question does not specify, "for what values of t are the two functions equal, and when the two functions are equal to y1 (and t=t1), then the equation with both functions on either side are equal, and is that not a solution to the problem, "at what value of y is the equation true?" The question asks for a solution, and does not specify it needs to be a t value.(5 votes)
- It's asking what values you could input into the two equations to make them equal, ergo produce the same output.

The y value is the dependent variable and the t value is the independent variable, therefore the t value is the input and the y value is the output.

If you stuck the y value into the equation as the t value then the results might not be the same as they may output different variables.

It's not asking you for what the answer to the equation is, it's asking you for what values you can input that will solve the problem, the problem being the need to know which input variables will produce the same output.(10 votes)

- Sal says the graph of h(t) is yellow at0:26, but it looks blue to me. Whats up?(7 votes)
- Maybe Sal meant blue but thought yellow. Also, blue and yellow make green. :D(3 votes)

- Why didn't Sal use the Y answers?(2 votes)
- Is "e^(2t) - 2t^2" classified as polynomial?(2 votes)
- As long as you have (e^2)t and not e^(2t) where "t" is in the exponent, then yes, it would be a polynomial.(1 vote)

- why did we only plot the t part (t1 and t4)

why not the y part ?(1 vote)- The question asks which values appear to be solutions.

𝑦₁, for example, is equal to 𝑓(𝑡₁),

and would solve the equation if 𝑓(𝑓(𝑡₁)) = ℎ(𝑓(𝑡₁)),

which we can't tell from the graph since we don't know how the axes are scaled, but it appears unlikely that 𝑦₁ would be a solution and therefore we don't count it as one.(1 vote)

- I don't understand why we have t sub 1 and not t equals 1.

What does this represent?(1 vote) - 101-102=1 which is wrong ?if i have to find the correct answer there are few conditions:

u should replace only one digit and place it some where else?the operations cannot be changed?(0 votes)- 101-102 = -1, not 1

Seems like you need to review adding/subtracting signed numbers (integers). Here is the link: https://www.khanacademy.org/math/algebra-home/pre-algebra/negatives-absolute-value-pre-alg#add-subtract-negatives-pre-alg

Try using a number line. It is a great tool to help you get the sign correct.

Start at zero. Go right to 101 on the number line.

To do -102, you would then move 102 units to the left (take away 102 units).

You will end up on -1. If you end up on +1, then you have only moved 100 units to left.(4 votes)

- Does this problem illustrate the pitfalls of mathematical when expressed in words? If I input a specific t value, I get a specific y value. When inputting t1, the same output, y1, occurs in both f(t) and y(t) equations. Inputting y1 or y4 produces no value for either equations and therefore not solutions... or are they? The question asks "which of the following appear to be solutions of ...." Is the input the solution or the output the solution? I am used to thinking of the output as the solution, and therefore y, unless you are telling me now that the value inserted into an equation is the solution and the not the value produced when operations are performed upon the input. That's one problem - the way the question is worded. A deeper problem is logical --- If I take the equation x=y-5 and express it as f(x)=y-5, you will tell me that y is a function of x. I disagree. x=y-5 is the same as y=x+5 and therefore f(y)=x+5. Isn't it silly to say x is a function of y or y is a function of x, when in fact each is a function of the other? Graphically, all I have to do is flip the page a quarter turn and I get the inverse relationship and can clearly see what you call the output is also the input of the function going in reverse. So when you tell me y is a function of operations performed on t (that you can map from t to y with the aid of logical/math operations), you also imply that you can map back from y to t with an inverted function. If my thinking is incorrect, please correct me.(0 votes)
- Which of the following appear to be solutions of
`a=b`

?`a`

,`b`

Let's try again. Assume`f(a)=b`

and`g(a)=b`

. Which of the following appear to be solutions of`f(a)=g(a)`

?`a`

,`b`

You mentioned - "The solution of an equation is always a value for the unknown".`f(t1)=e^(2t1)-2t1^2=y1`

`h(t1)=4-5t1^2=y1`

`4-5t1^2=e^(2t1)-2t1^2=y1`

From this point of view(intuitively), solution to the unknown(dependent) seems to be`y1`

, that is, given the input`t1`

what is the output.

Clearly, some functions can't be reversed while keep it consistent. But we've two completely different functions and by no mean I'm suggesting the view by reversing any functions.

Let me know what you think.(1 vote)

## Video 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.