for the last question put x+x+xsqrt2 = 94 +94sqrt2 on desmos and you'll get the right answer, but of course understand the algebraic way so that you don't rely on desmos entirely
The roots of f’(x) tell us when f(x) is at a minimum, maximum, or inflection point. This is because the slope of the tangent will be 0 at that point. Since g(x) is a quadratic, we know the roots of its derivative will tell us when it reaches its vertex.
You can take short cuts if you understand the principle. The formula makes sense because of the method shown in the video. If you fully understand then by all means use the formula!
A much faster way of finding the answer for the problem where you are trying to find the minimum value of g(x) is that you know -b/2a equals -8. Because we are looking for the answer when we plug in x+5 you just move -8 to the left 5 units which is -13. Much faster than fully factoring it out and can be the difference between finishing the test and not.
for the first question since the y intercept needs to have x=0 so you would know k=5 and then put into desmos the table 5,13 and 12,-15 and do ysub1 tilda mxsub1+b and look at the line....
For the purpose of the SAT exam, it is essential to get the answer as quickly as possible. In the second problem, f(x+5) is a horizontal shift operation. We can take the vertex of f(x), and then shift it 5 units to the left. The vertex of f(x) can be found as V(x) = -b/2a = -64/8 = -8 The vertex of f(x+5) is then equal to V(x+5) = -8 -5 = -13 If you cannot remember the shortcut formula, you can always complete the square. But do not complete the whole thing. Save time whenever you can. Completing the square, 4x^2 + 64x 4 (x^2 + 16x) 4 (x + 8)^2 The vertex is x = -8
@@stier5558 It looks like you are one of those who relies heavily on Desmos. Be sure you read the graph correctly. If you are not interested in conceptual analysis, that is always your privilege.
On the second problem did anyone else overcomplicate things and use calculus by expanding the g(x) function, taking the derivative, setting it equal to 0, and solving algebraically?
Creating g(x) and finding its minimum do not complicate things. That is what the question is asking for. You complicate if you are doing something that is not being asked.
In the last problem, we can apply the GUESS & CHECK approach to get the answer quickly. A. 47 + 47 + 47 rad(2) Wrong B. 47 rad(2) + 47 rad(2) + 47(2) CORRECT ANSWER C. 94 + 94 + 94 rad(2) Wrong D. 94 rad(2) + 94 rad (2) + 94(2) Wrong Once you get the correct answer, there is no need to test the other answer choices. Your answer is correct by definition. Again, save time whenever you can.
@@DrewWerbowski You’re welcome. My suggestion is you describe several methods (if the problem allows) and then provide details on the preferred method.
guys please use desmos for most of these question. it will save your time. this guy is explaining how to get to the answer without desmos. most of these question can be done with basic understandings and some lines on desmos.
@@raylvr5485 yes, as the comment said before, most of these questions can be solved using desmos and it saves so much time. But again, it shouldn't be relied on entirely
Desmos is a great tool if you have access and I’d encourage you to use it for time on the DSAT. That being said, ensure you understand how to do it without because you often won’t have access for midterm or final in college
you deserve more likes and subscriptions tbh
U do a great job
I am sharing it with all of my friends!
Thank you so much 😀 Good luck with your studies!
babe wake up drew werbowski uploaded again again
Yessss! Thank you for another one.
More to come!
For the first one the gradient is negative but you drew a positive slope
for the last question put x+x+xsqrt2 = 94 +94sqrt2 on desmos and you'll get the right answer, but of course understand the algebraic way so that you don't rely on desmos entirely
I did not get this: at 6:52, how can taking a derivative helps us find the roots of vertex?
The roots of f’(x) tell us when f(x) is at a minimum, maximum, or inflection point. This is because the slope of the tangent will be 0 at that point. Since g(x) is a quadratic, we know the roots of its derivative will tell us when it reaches its vertex.
for question #2 can't you used -b/2a to find the value of x because a parabola reaches it's minimum at it's vertex?
Yes you absolutely can, the method he used is just so complex for no reason. Using the formula -b/ 2a prevents anyone from any unnecessary mistake.
You can take short cuts if you understand the principle. The formula makes sense because of the method shown in the video. If you fully understand then by all means use the formula!
@@zedroxextrage1045Bruh, I just used guess and check
A much faster way of finding the answer for the problem where you are trying to find the minimum value of g(x) is that you know -b/2a equals -8. Because we are looking for the answer when we plug in x+5 you just move -8 to the left 5 units which is -13. Much faster than fully factoring it out and can be the difference between finishing the test and not.
Love this method! Thanks for commenting
These vids are really helpful
for the first question since the y intercept needs to have x=0 so you would know k=5 and then put into desmos the table 5,13 and 12,-15 and do ysub1 tilda mxsub1+b and look at the line....
For the purpose of the SAT exam, it is essential to get the answer as quickly as possible. In the second problem, f(x+5) is a horizontal shift operation. We can take the vertex of f(x), and then shift it 5 units to the left. The vertex of f(x) can be found as
V(x) = -b/2a = -64/8 = -8
The vertex of f(x+5) is then equal to
V(x+5) = -8 -5 = -13
If you cannot remember the shortcut formula, you can always complete the square. But do not complete the whole thing. Save time whenever you can. Completing the square,
4x^2 + 64x
4 (x^2 + 16x)
4 (x + 8)^2
The vertex is x = -8
you can also take derivative of f(x) and then put f'(x)=0. finding x then you get the answer
@@hamdanmalani9501
It is a good alternative if you know how to do derivatives.
f’(x) = 2ax + b = 0
From which you get
x = -b/2a
The vertex formula.
You don’t have to do work for this question at all, just plug x + 5 for all the xs into the desmos graphing calculator and youll have your minimum
@@stier5558
It looks like you are one of those who relies heavily on Desmos. Be sure you read the graph correctly. If you are not interested in conceptual analysis, that is always your privilege.
@@OverclockingCowboy You’re not wrong, I do heavily rely on it
On the second problem did anyone else overcomplicate things and use calculus by expanding the g(x) function, taking the derivative, setting it equal to 0, and solving algebraically?
Nothing wrong with this! It’s not the fastest way but shows your understanding. Probably would have only saved about 1-2 minutes
Creating g(x) and finding its minimum do not complicate things. That is what the question is asking for. You complicate if you are doing something that is not being asked.
For the second one we can derivate f(x) and then equate f’(x) with 0
Yes, you can do this too
In the last problem, we can apply the GUESS & CHECK approach to get the answer quickly.
A. 47 + 47 + 47 rad(2)
Wrong
B. 47 rad(2) + 47 rad(2) + 47(2)
CORRECT ANSWER
C. 94 + 94 + 94 rad(2)
Wrong
D. 94 rad(2) + 94 rad (2) + 94(2)
Wrong
Once you get the correct answer, there is no need to test the other answer choices. Your answer is correct by definition. Again, save time whenever you can.
Thanks for your feedback! These are great things to keep in mind. I’ll definitely address the fastest method of solving in future videos.
@@DrewWerbowski
You’re welcome. My suggestion is you describe several methods (if the problem allows) and then provide details on the preferred method.
guys please use desmos for most of these question. it will save your time. this guy is explaining how to get to the answer without desmos.
most of these question can be done with basic understandings and some lines on desmos.
Desmos is great to save time, but should not be relied on entirely. Building math understanding and problem solving ability will help in the long run!
@@DrewWerbowski agreed! just saying that if one have the understanding of what the question is asking they can use desmos to save time.
Can desmos be used in the exam?
@@raylvr5485 yes
@@raylvr5485 yes, as the comment said before, most of these questions can be solved using desmos and it saves so much time. But again, it shouldn't be relied on entirely
Please make more videos on SAT
Will do!
for the last question, when i plug in 47(sqrt)2 into 2L+L(Sqrt)2 i get 94+47(Sqrt)2, not what the question wants. Any help?
Sounds like an algebra mistake. Try to follow along with the video step by step and see where you get a different result
@@DrewWerbowski aahh youre right 😭I got it now, thanks!!
Q2 put in desmos u get answer -13
Can u use calculator for the last question
Yes, if you have a calculator that supports simplified radicals then you could input your expression for L.
@DrewWerbowski Thank u so much 💓
These questions are damn easy
the second question is a 10 second job in desmos lmfao all u have to do is input f(x) 4x^2 + 64x + 262 and g(x) = f(x +5) and its a free question
Desmos is a great tool if you have access and I’d encourage you to use it for time on the DSAT. That being said, ensure you understand how to do it without because you often won’t have access for midterm or final in college
@@DrewWerbowski yeah ofc i get the problem its just way easier to use desmos in the context of the sat. appreciate your videos by the way!