Medal to helper! Can you please check my answers? Important! 5Q's. Thanks!

- anonymous

Medal to helper! Can you please check my answers? Important! 5Q's. Thanks!

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- katieb

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- anonymous

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

- anonymous

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

- anonymous

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)

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- anonymous

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

##### 1 Attachment

- anonymous

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

- anonymous

@phi @Hero @SolomonZelman @tom982

- SolomonZelman

#1 Correct
#2 Correct
#3 Wrong
#4 Wrong
#5 Wrong

- SolomonZelman

No, #3 might be right, let me check again

- SolomonZelman

(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

- SolomonZelman

yes, #3 is correct, but #4 and #5 are for sure wrong.

- anonymous

Okay thanks. Could you help me with those two?

- SolomonZelman

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

- SolomonZelman

So, your coefficient must be negative, because the parabola opens down.

- SolomonZelman

And, tell me where is the center of your parabola located?

- SolomonZelman

I mean the vertex, not the center

- anonymous

(3,2)

- SolomonZelman

Yes, that is right.

- anonymous

So, y = − (x − 3)^2 + 2 is correct, right?

- SolomonZelman

\(\large\color{#000000 }{ \displaystyle y=a(x-h)^2+k }\)
h=3
k=2
\(\large\color{#000000 }{ \displaystyle y=a(x-3)^2+2 }\)
So the closes option is
\(\large\color{#000000 }{ \displaystyle y=-(x-3)^2+2 }\)

- anonymous

Alright thank you. Now number 5.

- SolomonZelman

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?

- anonymous

I would like to move on to #5.

- SolomonZelman

Alright ...

- SolomonZelman

You have to expand:
\(\large\color{#000000 }{ \displaystyle (x-4)(x^2+3x+2) }\)

- SolomonZelman

Can you tell me what do you get if you were to expand
\(\large\color{#000000 }{ \displaystyle (x-4)\cdot x^2 }\)

- SolomonZelman

You can use the following rules
\(\large\color{#000000 }{ \displaystyle (a+b)\times c=(a\times c)+ (b\times c) }\)
\(\large\color{#000000 }{ \displaystyle (a-b)\times c=(a\times c)- (b\times c) }\)

- anonymous

multiply x^2 and x, then multiply x^2 and -4?

- SolomonZelman

yes, and wha do you get after doing this?

- anonymous

x^3 - 4x^2

- SolomonZelman

yes, very good

- SolomonZelman

Now, what do you get after expanding
\(\large\color{#000000 }{ \displaystyle (x-4)\times 3x }\)

- anonymous

3x^2 - 12x

- SolomonZelman

Yes, very good

- SolomonZelman

And lastly what do you get after expanding
\(\large\color{#000000 }{ \displaystyle (x-4)\times 2 }\)

- anonymous

2x - 8

- SolomonZelman

Yup, now we will put the peaces together...

- SolomonZelman

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:
\(\large\color{#000000 }{ \displaystyle x^3-4x^2+3x^2-12x+2x-8 }\)

- SolomonZelman

Can you simplify
\(\large\color{#000000 }{ \displaystyle x^3-4x^2+3x^2-12x+2x-8 }\)
?

- anonymous

Yes, let me try.

- SolomonZelman

go ahead and take your time ;)

- anonymous

x^3 - x^2 - 10x - 8

- SolomonZelman

yup, and that is your final answer.

- anonymous

Wow. I would have never guessed that one. You're explanation makes sense. Thank you so much!

- anonymous

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

- SolomonZelman

|dw:1449680605870:dw|

- anonymous

Gotcha.

- SolomonZelman

You number is:
2-8i
So the conjugate is
2+8i (YES, YOU ARE CORRECT)
(or in fact it could also be -2-8i, but THAT would be a negative conjugate)

- anonymous

Coolio. This one is giving me trouble as well. Simplify the expression |dw:1449680910058:dw|

- anonymous

Answer Choices: (Just for reference)
A. \[\frac{ 12-9i }{ 25 }\]
B. \[\frac{ 12-9i }{ 7 }\]
C. \[\frac{ 9+12i }{ 25 }\]
D. \[\frac{ 9+12i }{ 7 }\]

- anonymous

I chose C.

- SolomonZelman

You are probably aware of these two facts:
\(\large\color{#000000 }{ \displaystyle (a-b)(a+b)=a^2-b^2 }\)
and, \(\large\color{#000000 }{ \displaystyle i^2=-1 }\)
So, the conjugate is there to eliminate the i.
So if you have \(\large\color{#000000 }{ \displaystyle (2i+4) }\), then as you multiply
that times the conjugate (in this case (2i-4), you get:
\(\large\color{#000000 }{ \displaystyle(2i+4)(2i-4)=(2i)^2-4^2=2^2\cdot i^2-16 }\)
\(\large\color{#000000 }{ \displaystyle=4\cdot (-1)-16 =-4-16=-20 }\)
But, why do we need the CONJUGATE tho, if we can't just randomly multiply every imaginary number times its conjugate (because you are unjustly changing the value).
Suppose you had the following problem.
\(\large\color{#000000 }{ \displaystyle \frac{i+1}{2i+4} }\)
and you want to rationalize the denominator
(remove any roots from the denominator)
Then, you would multiply top and bottom
by the conjugate of 2i+4 (which is 2i-4)
\(\large\color{#000000 }{ \displaystyle \frac{(i+1)\color{red}{\times (2i-4)}}{(2i+4)\color{red}{\times (2i-4)}} }\)
You already know that on the bottom you will get a -20.
\(\large\color{#000000 }{ \displaystyle \frac{(i+1)\color{red}{\times (2i-4)}}{-20} }\)
and as we simplify the to we get:
\(\large\color{#000000 }{ \displaystyle (i+1)\times (2i-4) }\)
\(\large\color{#000000 }{ \displaystyle (i+1)\times 2i=2i^2+2i=2(-1)+2i=-2+2i }\)
\(\large\color{#000000 }{ \displaystyle (i+1)\times 1=i+1 }\)
add the results, you get:
\(\large\color{#000000 }{ =-2+2i +i+1=-1+3i \quad or,~=3i-1 }\)
So \(\large\color{#000000 }{ \displaystyle (i+1)\times (2i-4)=3i-1 }\)
And thus you get:
\(\color{#000000 }{ \displaystyle\frac{i+1}{2i+4}= \frac{(i+1)\color{red}{\times (2i-4)}}{(2i+4)\color{red}{\times (2i-4)}}=\frac{3i-1}{-20} }\)
which can also be written as (1-3i)/20.

- SolomonZelman

In case of you problem, you have to add like terms in the denominator first, and then use my example to rationalize the denominator.

- SolomonZelman

One note for the NUMERATOR, though,
\(\large\color{#000000 }{ \displaystyle \sqrt{-9}=\sqrt{(-1)}\cdot \sqrt{9}=i\times 3=3i }\)

- anonymous

That's a lot of information. Thanks.

- SolomonZelman

I will provide the steps, just in case:
{ Step 1 }
Your denominator is (3-2i)+(1+5i).
Simplify the denominator.
{ Step 2 }
Multiply top and bottom times the
conjugate of the new denominator.

- SolomonZelman

Note, for step 2, apply,
(a-b)(a+b)=a²-b²
just as I did, and then simplify.
Also don't forget that i²=-1

- anonymous

I got this answer this time: \[\frac{ 9+12i }{ 7 }\]

- anonymous

Is this correct? @SolomonZelman

- SolomonZelman

you made a little error
(4+3i)(4-3i) = is NOT 16-9
RATHER,
(4+3i)(4-3i) = is going to be = 16 + 9

- SolomonZelman

because
3i × (-3i)= -3i² = -3(-1) = 3.

- anonymous

Yay! I got a 100%! Thank you so much for your time, patience, and help! @SolomonZelman

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