October 2024 SAT Prep: You MUST Know This HARD Problem

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You MUST Know This HARD Problem
In this video, Darren develops a solution for a Quadratic problem from a Bluebook practice exam. This problem utilizes key concepts and formulas that will 100% guaranteed be tested on future SAT administrations.
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It is important to know some of the most important tips, tricks, and strategies for the new Digital SAT Math section. You will get practice with the Digital SAT Math questions testing your skills on quadratics, exponential functions, exponents, lines, special quadratics, and much more.
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Опубликовано:

3 окт 2024

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Комментарии : 31

@noth1ngnss921 Месяц назад

There is another way to solve this problem. Since f(x) is a quadratic function, and f(-9) = f(3), we know that f(x) must be symmetrical across x = -3. That means the vertex is at (-3, k). Plugging this into the vertex form of the function, we get: a(x + 3)^2 + k. Converting that back into the standard form, we get: ax^2 + 6ax + 9a + k. We now know that 6a = 4, thus a = 2/3, proving statement II to be false. We also know that 9a + k = c, or 6 + k = c. So c < 6, making statement I not true either.

@PrepworksEducation Месяц назад

🙌

@hehehe3837 Месяц назад

We can also solve this question by desmos. Just plug in the f(x), f(-9) and f(3) and after that use the slider of a and c and check if the conditions shown is true or not.

@PrepworksEducation Месяц назад

Technically so, there is some possibility for error however.

@OverclockingCowboy Месяц назад

@@PrepworksEducation Technically YES and NO. If you know what you are doing, you would look for the correct value of “a” first. Once you have the correct value for “a”, then you evaluate “c” under the condition k < 0. Otherwise, you would just be sliding a and c the whole day. If you know what you are doing, you would not even attempt to use Desmos for this problem. So, the first exercise is: How do you find “a” in Desmos? We know that a = 2/3.

@PrepworksEducation Месяц назад

True. Lots of ways to solve the problem, as per usual.

@OpokuMarvin-wx8ei 13 дней назад

Pls since “k” is negative, am I right to use the discriminant b^2-(4ac)>0? The graph of the function will cross the X-axis. Here, “b”is 4, “a” is 2/3 and “c” is c 16-4(0.667-c)>0 c

@AdamHyink 2 месяца назад

For verifying the second postulate, could you also use the fact that the midpoint between -9 and 3 must be the x-value of the vertex, and from there, use -b/2a to calculate the value of a?

@PrepworksEducation 2 месяца назад

If you were able to determine the “b” value, then yes - that would be a great alternative!

@apsidalsolid4607 Месяц назад

@@PrepworksEducation I believe the b value is given, as "4x" is given in the equation. Therefore, the b value would just be 4. Am i correct in thinking so?

@PrepworksEducation Месяц назад

@@apsidalsolid4607 The coefficient in front of a term would dictate its numerical value, yes.

@OverclockingCowboy Месяц назад

You definitely can determine “a” from -b/2a. -3 = -4/2a a = 2/3 The author is using a = 48/72 that can be simplified as 2/3.

@PrepworksEducation Месяц назад

Indeed!

@Chilldude-zh1lt 2 месяца назад

Is this calculus? Im taking precal next year but this looks strikingly similar to algabra two class! Also thank you so much for making these! I appreciate it a lot!

@PrepworksEducation 2 месяца назад

No problem! All the math content on the SAT is algebra 2 or below.

@E-flat 27 дней назад

@@Chilldude-zh1lt No, there is no calculus on the SAT.

@rky824 2 месяца назад

I just did a system of equations between f(-9)=0 and f(3)=0, I still get: "neither I nor II" but I'm not sure if the method makes sense.

@PrepworksEducation 2 месяца назад

Hey! I know that you can solve this problem using desmos + sliders via the regression feature. Could you run us through your method further?

@OverclockingCowboy Месяц назад

@@PrepworksEducation You can abuse Desmos on this problem in the same manner that you solved the problem in the video. Calculate “a” in regressing the standard form by using f(-9) = f(3). You may have to set c to a known value, eg, 0 or 1. Note that the value of “a” is independent of c. So any value of c would give you the correct value of “a”. Substitute a = 2/3 in the standard form and put a slider on c. Slide c (y intercept) under the condition k < 0 (y component of vertex).

@darkotaku7456 Месяц назад

Hello, I solved c < 0 in a different way, now we know f(-9) = f(3), and we found a =2/3, cant we then substitute back to f(-9) = f(3) to get y? after we do so we get 18 and then we are left with 18 = 18+c(using either -9 or 3) then we get c =0 which doesn't satisfy the inequality so its wrong. Is this a wrong method?

@PrepworksEducation Месяц назад

Hey, unfortunately would not be the right approach. The problem here is that initially we used that equation to solve for “a”, we cancelled “c” on both sides - because “c” = “c”. By setting them equal to each other and finding the “a” value, we want them to be equivalent. This means that we substitute “a” = 2/3 into f(-9) = f(3), we are left with: 18 + c = 18 + c So, we can’t actually determine what the “c” value here is since any value would satisfy the equation.

@darkotaku7456 Месяц назад

@@PrepworksEducation That makes sense. Appreciated!!!

@PrepworksEducation Месяц назад

@@darkotaku7456 No problem!

@fedal5956 Месяц назад

How do we know c is negative?

@PrepworksEducation Месяц назад

Hey, it is one of the constraints given that we have to prove right/wrong. In this case, “c” doesn’t actually have to be negative.

@lehoangvu3540 2 месяца назад

To substantiate that fact that first condition (c (-9,0) substitute in your equation of (48/72)x^2+4x+c = 0. That way if c =18 then it cannot be

@PrepworksEducation 2 месяца назад

Hey, yes that works!

@OverclockingCowboy Месяц назад

Not really. f(-9) = f(3), but f(-9) is not equal to zero. f(-9) is not defined until you have a definite value for c. You need to evaluate c < 0 under the condition k < 0. That relationship is obtained in the video as c - 6 = k Since k < 0, the maximum value for c is 6. C is the y intercept, whereas k is the y component of the vertex.

@PrepworksEducation Месяц назад

Exactly!