f(x) = ax^2 + 4x + c; In the given quadratic function, a and c are constants. The graph of....
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- čas přidán 20. 08. 2024
- Bluebook Digital SAT Test 2 Module 2 (Hard) Question 22:
f(x) = ax^2 + 4x + c
In the given quadratic function, a and c are constants. The graph of y = f(x) in the xy-plane is a parabola that opens upward and has a vertex at the point (h, k), where h and k are constants. If k is less than 0 and f(-9) = f(3), which of the following must be true?
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This makes so much more sense than the convoluted answer they gave on Bluebook, and it matches my test taking style so much more. Thank you!!!!
You’re very welcome!!
This problem is easier to visualize with actual figures (number).
We know that the vertex of the parabola is at f(-3) [midpoint of f(-9) and f(3)]. We can apply this point into the vertex form.
a (x + 3) ^2 + k
ax^2 + 6ax + 9a + k
Equating coefficients with the standard form
b = 6a
c = 9a + k
From the first relation, replacing b by 4
4 = 6a
a = 2/3
Thus, condition II is false.
From the second relation,
c = 9a + k = 6 + k
Thus,
c > 0, in the interval -6 < k < 0
c < 0, when k < -6
Hence,
Condition I is false. c is not ALWAYS less than zero when k is less than zero.
What it means:
k = height of vertex
c = y intercept.
When the vertex of the parabola is touching the x axis at k = 0, the y intercept of the parabola is at c = 6.
When the vertex is at k = -6, the y intercept is at c = 0.
This is the range where c is greater than zero when k is less than zero.
Thanks
@@TheSATMathGuy
An interesting follow up to this question is: when is condition II true? What is the condition for a to be greater than 1?
It was derived above that b = 6a. Thus, if b > 6, then a > 1.
There is a general relationship between a and b, and the vertex V (x component) from the vertex equation.
V = -b/2a
Or
a = -b/2V
Condition II would have been a trap answer if b were given as 7, that is, the given equation was f(x) = ax^2 + 7x + c.
@@OverclockingCowboy 🙏🏾
@TheSATMathGuy Thank you so much for this video!
But I'm so sorry, even after reading the comment section, I'm still not understanding why c
Thanks for your question. c could be less than 0. It just doesn't have to be and the question asks: "which of the following MUST be true?". Secondly, f(-9) = f(3) can definitely happen where c > 0 or c < 0. For example, you can have points (-9, 5) and (3, 5). You are right that the vertex is at (-3, ?). The only place where you're slightly off is on how the given information affects the y-intercept. It could be negative or it could be positive.... we just really do not know for sure based upon any of the given information. Hope that helps
@@TheSATMathGuy Completely forgot to reply to this, but thank you sooo much! Your explanation and giving the (-9, 5) and (3, 5) example really helped. Thank you again!
@@NKA70 you’re very welcome!!
@@NKA70 but they said k
@@editer3089
(h, k) is the vertex which in this case is the lowest coordinate on the parabola. K doesn’t represent the y-intercept, it just represents the lowest y-coordinate the parabola can go.
What helped me realize that the y-intercept can be positive is to plug in the coordinates that “TheSATMathGuy” wrote. Try plugging in (-9,5) then a (3,5) with a vertex at (-3,-0.5) on a desmos graphing calculator.
If you imagine a parabola going through those three points, you can see that the y-intercept would be a positive number. Thus, c (the y-intercept) does not HAVE to be less than 0
Hopefully that helped you bit!
thanks for your videos, they've been super helpful! I was just wondering though, how is there a possibility that C (or the y-intercept) can be a positive value if we know that the parabola faces upwards and that they're two points across the x axis? Or in other words, if the vertex's Y value was greater than 0 how would it cross the x-axis still even when the parabola is facing upwards?
Thank you so much for watching the videos. I’m grateful that you found the channel and that it has been helpful. To answer your question: if the y value of the vertex of an upward facing parabola is positive (greater than 0) it CANNOT have any x-intercepts. If, however, the y value of the vertex is negative and faces upward - it will definitely have two x-intercepts and either a positive or negative y-intercept. You’d need more information in order to know exactly where the y-intercept is. Hope this helps.
@@TheSATMathGuy Oh that makes sense, I just had a brainfart because I was thinking that the C value in the quadratic equations was the y value in the vertex. Anyways thanks for the clarification and all your help!
@@zuzuotb1121 Any time!!
Thank you so much. Your videos help a lot! 😊
You’re very welcome!!
thank you so much for these! they help a lot
You’re welcome!
If k
not at all. The parabola can face upward with a vertex that is below the x-axis and a y-intercept that is positive. Hope that answers your question. I explain this in detail starting around the 2:40 mark.
K is the y point of the vertex while C is the point of the y intercept. So the vertex can be low but as the parabola curves up, it can still hit the y intercept(c).
thank you so much! I was super confused on this one
You’re welcome
why c is equal to y intercept?
The y-intercept is found by setting x equal to 0. When x = 0, f(x) = c. This means that c is the y-intercept. Hope that helps
is there any other way we can solve this? i do not get it
Yes - use f(-9) = f(3) to plug in -9 into the function for x and set it equal to plugging 3 in to the function for x. You'd get: 81a - 36 + c = 9a + 12 + c. Solve for an and you'll see that a = 2/3. Hope that helps!
@@TheSATMathGuy okay i see, when we write the equation we will get red of c right? which means it has no solution, i am not sure about this correct me please, and when a=2/3 it means that a is greater then ` not equal that is why it is wrong? is this go it is going
is the vertex and y intercept not the same thing?
Vertex is where the parabola is, and y intercept is where the parabola CROSSES the y
Vertex is the lowest point of an upward facing parabola or the highest point of a downward facing parabola. The y intercept is where the parabola intersects the y axis.
i dont get it if K is lesss than zero than should be y intercept less than zero
go to desmos and graph x^2+11x+28, you'll see that the vertex (h,k) is below the x-axis, but c=28, which is positive
the parabola has just been moved far enough to the left that the y intercept can be positive. I just translated f(x)=x^2+x+2 to the left by 5.
I wouldn't say "should" be. More like "could" be. The question asks which answer "must" be. "MUST" be means that it CANNOT be anything else. Do you feel that strongly about the y-intercept being less than zero?
@@TheSATMathGuy got it gave my sat two days back your videos helped alot
cheers mate 😉