Math unit test.. i need help xD

[paksjohn 0]

Alright so here's the series of questions that i don't wanna answer.. cuz i suck at math hard.. or maybe  i just hate math ok so here we go

1.The vertex of y=(x-3)² is?

2. If the vertex of the graph of y=10x+c is at the origin, what can you conclude about c?

3. What is the domain at the function y= 4-3x²

4. Describe the roots of x²+4x+7 = 0 (tell if it's imaginary, real or what so ever)

5. find the value of k such that f(x) = kx(x+4)-1 has exactly one zero

6. find the maximum value of the function y = -3x + 6x - 8

7. find the quadratic function whose zeros are 5 and 3

8. what is the value of the constant k such that x²-4x+2k = 0 has no real roots?

9. The graph of the function y = x²-4 passes throught (3,5). What are the coordinates of it's reflection across y-axis?

10. Find the quadratic function f(x) whose graph has its vertex at (-2,3) and contains the point (4,12)

 

[Crius 0]

1) The vertex of a function is the high and/or low points of the curve. The vertex of a function f(x)  is where the first derivative of the function f'(x) = 0. To make it easier to get the derivative, you might want to expand the function to

f(x) = x² - 6x + 9

Getting the derivative is pretty easy for this type of function where no variables are multiplied with each other. If you have ax^b, then the derivative is a*b*x^(b-1). The derivative of a constant is 0. Do that for all three parts of your function. From that you get the value of x at the vertex point. You can then find the value of y at the vertex by inserting the x value into the original function f(x).

 

2) You sure you copied that correctly? Unless there's something lost in translation, that function is a straight line and shouldn't have a vertex. For it to even intersect origo, c would have to be 0 though.

 

3) The domain of a function f(x) is the set of input values (x) that produce a valid output (f(x), or y). Are there any numbers for this function that would not work? If all numbers are valid, then you typically write ]-inf, inf[ as the domain.

 

4) If you have an equation x² + bx + c = 0, the roots are the x values where the equation is balanced. You can find these out by either factoring the equation into (x + n)(x + m) = 0 form (not possible in this case), and then -n and -m are the roots, or you can use the quadratic formula, which is (in the simplified form when a = 1):

x₁ = -b/2 + Sqrt((-b/2)² - c)

x₂ = -b/2 - Sqrt((-b/2)² - c).

An easy way to see if the roots are imaginary is to check if the part inside the square root is negative. If so, it's imaginary, otherwise it's real.

 

5) If I'm interpreting the question correctly, the trick here is to ensure that both roots have the same value. If we expand it we get

f(x) = kx² + 4kx - 1

To find the roots we set f(x) = 0, and in order to get exactly one root, we use the quadratic formula and get:

x₁ = (-4k + Sqrt((4k)² + 4k) / 2k

x₂ = (-4k - Sqrt((4k)² + 4k) / 2k

From this we can tell that k is definitely not 0 since then we'd get a division by 0 which is bad. The only way that both roots could be the same is if the content of the square root expression is 0. So, we have

(4k)² + 4k = 0

4k * (4k + 1) = 0

or with factoring:

(4k + 0)(4k + 1) = 0

So either k can be 0, or it can be -1/4th. But since we've already determined that k cannot be 0, it must be 1/4th. We can test this by inserting it into the original function:

-(1/4)x(x+4)-1 = 0

-(1/4)x² - x - 1 = 0

x² + 4x + 4 = 0

x₁ = 2 + Sqrt(4 - 4)

x₂ = 2 - Sqrt(4 - 4)

Yep, they are the same.

 

6) Again I suspect you didn't copy it correctly (seems like there's a power missing in there somewhere). Anyway, finding the maximum value can be done by using the same method as in #1. Get the derivative of the function, check for which x value the derivative is 0, and then insert the x value into the original function to get the y value (which is the maximum or minimum value of the function, depending on how it is shaped).

 

7) Well, with factoring, this is trivial. Remember (x + n)(x + m) = 0 from #4? Replace n and m with your roots, and that's your function.

 

8) Again we can use the quadratic formula and get

x₁ = 2 + Sqrt(4 - 2k)

x₂ = 2 - Sqrt(4 -2k)

For a non-real root, the square root expression must be negative. In this case, any number which satisfies 4 -2k < 0, or 2 - k < 0 works.

 

9) I'm not sure I understand the question, but the function is symmetric around the y-axis. In other words, -x and +x gives you the same y value, or in this case, two points (3,5) and (-3,5).

 

10) The vertex at point (-2,3) means that f'(-2) = 0 and f(-2) = 3. We also have that f(4) = 12 since we know the function passes through that point. It's a quadratic function, so it will have the form f(x) = ax² + bx + c and thus f'(x) = 2ax + b.

Okay, so let's take this apart then. We have the following:

#1) -4a + b = 0 (from f'(-2) = 0)

#2) 16a + 4b + c = 12 (from f(4) = 12)

#3) 4a - 2b + c = 3 (from f(-2) = 3)

We can start by replacing b, so that b = 4a (from #1) and insert into the second two.

#2) 16a + 16a + c = 12

#3) 4a - 8a + c = 3

Next we replace c, with c = 12 - 32a (from #2) and insert it into #3

4a - 8a + 12 - 32a = 3

36a = 9 -> a = 1/4

Now we know what a is, so we can find out what b is from #1 as well

-4*1/4 + b = 0 -> b = 1

Finally, we insert this into #2 to get c

16*1/4 + 4*1 + c = 12 -> c = 4

 

Okay, so we know f(x) = ax2 + bx + c, and that a = 1/4, b = 1 and c = 4. So...

f(x) = x²/4 + x + 4

 

We can check this function to see if it satisfies the initial conditions. I won't, but it does.

[Roba 0]

When I look at all this, I get so glad for being in law school.

 

GOODBYE MATH HAHAHHAHAHAHAAHHAHAHAHAHAHAHAHHAHAA

 

 

I'm not mad

[Crius 0]

When I look at all this, I get so glad for being in law school.

 

GOODBYE MATH HAHAHHAHAHAHAAHHAHAHAHAHAHAHAHHAHAA

 

 

I'm not mad

 

I'm glad I'm not in law school pretty much all the time.

[Roba 0]

When I look at all this, I get so glad for being in law school.

 

GOODBYE MATH HAHAHHAHAHAHAAHHAHAHAHAHAHAHAHHAHAA

 

 

I'm not mad

 

I'm glad I'm not in law school pretty much all the time.

 

Didn't get it  ???

[Crius 0]

I don't like law, basically.

[-assKICKER- 1]

Engineering FTW .

[Liek 0]

omfg so hard ???

[mafia11 0]

i would have the answers for some of them, but our friends there already have the answers for all of them... im brazilian moved to united states, im a sophomore taking algebra 2, i have learn stuffs about that... pretty sure im going to learn all of it...

[Gogg 3]

When I look at all this, I get so glad for being in law school.

 

GOODBYE MATH HAHAHHAHAHAHAAHHAHAHAHAHAHAHAHHAHAA

 

 

I'm not mad

+1

 

Scary numbers!

[RauL~ 1]

looks cool.. i like math.. in Economy have a lot too

[paksjohn 0]

LOL. i need exact answer. not some guidelines on how to solve that problem. i hate math -.- lol still on going.. gogogo answer it now!

[Gogg 3]

What about do your stuff yourself?

Could've said "thanks" at least

 

[~Trinitas~ 0]

LOL. i need exact answer. not some guidelines on how to solve that problem. i hate math -.- lol still on going.. gogogo answer it now!

an exact answer huh

 

so in otherwords someone to do your homework for you :3 

rename your thread D8<

[Liek 0]

LOL. i need exact answer. not some guidelines on how to solve that problem. i hate math -.- lol still on going.. gogogo answer it now!

an exact answer huh

 

so in otherwords someone to do your homework for you :3 

rename your thread D8<

 

lolol

[YourMama 0]

Nerds itt

 

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