November 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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[nothngnss]

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.

[AdamHyink]

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?

[OpokuMarvin-wxei]

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<6.00 Thus c must not necessarily be less than 0

[hehehe]

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.

[darkotaku]

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?

[lehoangvu]

To substantiate that fact that first condition (c<0) is wrong, am I right to work out c = -18? I use f(-9)=0 => (-9, 0) substitute in your equation of (48/72)x^2+4x+c = 0. That way if c =18 then it cannot be <0?

[subba]

How is c<0 wrong? Doesn't c<0 fall in the domain of c<6? So even though the expression is not the same all the numbers that work for c<0 will make the equation true. Doesn't that mean it must be true?

[rky]

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.