Medal to helper! Can you please check my answers? Important! 5Q's. Thanks!
1. What is the slope of the line which passes through (−2, 0) and (0, 4)? A. −2 B. 0 C. 2 <--- My Answer D. Undefined
2. Use the table below to answer this question: x y −1 7 3 3 5 1 Find the average rate of change for the given function from x = −1 to x = 5. A. −6 B. −1 <--- My Answer C. 1 D. 6
3. What is the equation in point−slope form of the line passing through (0, 6) and (1, 3)? A. (y − 3) = −3(x − 1) <--- My Answer B. (y + 3) = 3(x + 1) C. (y + 3) = −3(x + 1) D. (y − 3) = 3(x + 6)
4. What is the equation of the graph below? A. y = − (x + 2)^2 + 2 B. y = − (x − 3)^2 + 2 C. y = (x − 2)^2 + 2 D. y = (x + 3)^2 + 2 <--- My Answer
5. Simplify (x − 4)(x^2 + 3x + 2). A. x^3 − x^2 + 14x − 8 B. x^3 + 7x^2 − 10x − 8 C. x^3 − x^2 − 10x − 8 D. x^3 + 7x^2 + 14x − 8 <--- My Answer
@phi @Hero @SolomonZelman @tom982
#1 Correct #2 Correct #3 Wrong #4 Wrong #5 Wrong
No, #3 might be right, let me check again
(0, 6) and (1, 3) slope=m=(6-3)/(0-1)=3/(-1)=-3 So, with slope -3 and point (1,3) you have: y - 3 = -3•(x-1) So #3 is correct. Alternatively, you could have used the other point (0,6) y - 6 = -3•(x-0) y - 6 = -3x
yes, #3 is correct, but #4 and #5 are for sure wrong.
Okay thanks. Could you help me with those two?
Yes, surely. the general form of the equation of the parabola with vertex at (h,k) is: \(\large\color{#000000 }{ \displaystyle y=a(x-h)^2+k }\) If the parabola opens up, a>0 if the parabola opens down, a<0
So, your coefficient must be negative, because the parabola opens down.
And, tell me where is the center of your parabola located?
I mean the vertex, not the center
(3,2)
Yes, that is right.
So, y = − (x − 3)^2 + 2 is correct, right?
\(\large\color{#000000 }{ \displaystyle y=a(x-h)^2+k }\) h=3 k=2 \(\large\color{#000000 }{ \displaystyle y=a(x-3)^2+2 }\) So the closes option is \(\large\color{#000000 }{ \displaystyle y=-(x-3)^2+2 }\)
Alright thank you. Now number 5.
If you want, also, you can solve for "a", and get the answer without eliminating the rest of the options. Want to solve for a, or to proceed to #5 right away?
I would like to move on to #5.
Alright ...
You have to expand: \(\large\color{#000000 }{ \displaystyle (x-4)(x^2+3x+2) }\)
Can you tell me what do you get if you were to expand \(\large\color{#000000 }{ \displaystyle (x-4)\cdot x^2 }\)
You can use the following rules \(\large\color{#000000 }{ \displaystyle (a+b)\times c=(a\times c)+ (b\times c) }\) \(\large\color{#000000 }{ \displaystyle (a-b)\times c=(a\times c)- (b\times c) }\)
multiply x^2 and x, then multiply x^2 and -4?
yes, and wha do you get after doing this?
x^3 - 4x^2
yes, very good
Now, what do you get after expanding \(\large\color{#000000 }{ \displaystyle (x-4)\times 3x }\)
3x^2 - 12x
Yes, very good
And lastly what do you get after expanding \(\large\color{#000000 }{ \displaystyle (x-4)\times 2 }\)
2x - 8
Yup, now we will put the peaces together...
We wanted to expand \(\large\color{#000000 }{ \displaystyle (x-4)(x^2+3x+2) }\) And this is the same as expanding \(\large\color{#000000 }{ \displaystyle (x-4)\cdot x^2 }\) \(\large\color{#000000 }{ \displaystyle (x-4)\cdot3x }\) \(\large\color{#000000 }{ \displaystyle (x-4)\cdot2 }\) separately, and then adding the results. We get that: \(\large\color{#000000 }{ \displaystyle (x-4)\cdot x^2=\quad +x^3-4x^2 }\) \(\large\color{#000000 }{ \displaystyle (x-4)\cdot3x=\quad +3x^2-12x }\) \(\large\color{#000000 }{ \displaystyle (x-4)\cdot2 =~{{\tiny~~}}\quad +2x-8}\) And add the results: \(\large\color{#000000 }{ \displaystyle x^3-4x^2+3x^2-12x+2x-8 }\)
Can you simplify \(\large\color{#000000 }{ \displaystyle x^3-4x^2+3x^2-12x+2x-8 }\) ?
Yes, let me try.
go ahead and take your time ;)
x^3 - x^2 - 10x - 8
yup, and that is your final answer.
Wow. I would have never guessed that one. You're explanation makes sense. Thank you so much!
Could you check this too?: Which of the following is the conjugate of a complex number with 2 as the real part and −8 as the imaginary part? A. −2 + 8i B. 2 + 8i <--- My Answer C. 2 − 8i D. −2 − 8i
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