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thank you so much and sorry for the typo its not "FTH'' ITS ''FIFTH''

How about entering your question properly.

You can edit your question I think, otherwise you can repost and I will delete your old one.  

Are u able to help me?

\(f(x) = \dfrac 1 3 x^2 - 3x + 12 = \\ \dfrac 1 3(x^2 -9x+36 )= \\ \dfrac 1 3\left(\left(x-\dfrac 9 2\right)^2-\dfrac{81}{4}+36\right) =\\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2- \dfrac{27}{4}+12 = \\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2+\dfrac{21}{4}\)

\(f(x)=0\\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2 = -\dfrac{21}{4}\\ \left(x-\dfrac 9 2\right)^2= -\dfrac{63}{4}\\ x-\dfrac 9 2 = \pm i\sqrt{\dfrac{63}{4}}\\ x = \dfrac 9 2 \pm i\dfrac 3 2 \sqrt{7}\\ x = \dfrac 3 2\left(3\pm i \sqrt{7}\right) \)

 Trigonometric Identities:

\(\large\dfrac{2\tan(x)}{1-\tan^2(x)} = 1\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{2\tan(x)}{1-\tan^2(x)} } \\\\ &=& \dfrac{ \tan(x)+\tan(x)}{1-\tan(x)\tan(x)} \quad | \quad \boxed{ \text{Formula: } \tan(x+y) = \dfrac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} } \\\\ &\mathbf{=}& \mathbf{\tan(2x)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \tan(2x) &=& 1 \\ 2x &=& \arctan(1) + n\pi,\ n \in \mathbb{Z} \quad | \quad \arctan(1) = \dfrac{\pi}{4} \\ 2x &=& \dfrac{\pi}{4} + n\pi \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{\pi}{8} + n\dfrac{\pi}{2}} \\ \hline \end{array}\)

\(\text{Split } \mathbf{\dfrac{\pi}{8} + n\dfrac{\pi}{2}}\text{ in two solutions:}\)

\(\begin{array}{|rcll|} \hline x & = & \dfrac{\pi}{8} + n\dfrac{\pi}{2} \\ x_1 &=& \dfrac{\pi}{8} + n\left(\dfrac{\pi}{2}+\dfrac{\pi}{2} \right) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{\pi}{8} + n\pi} \\\\ x_2 &=& \dfrac{\pi}{8}+\dfrac{\pi}{2} + n\left(\dfrac{\pi}{2}+\dfrac{\pi}{2} \right) \\ \mathbf{x_2} &\mathbf{=}& \mathbf{\dfrac{5\pi}{8} + n\pi} \\ \hline \end{array}\)

\(\dfrac{1}{|m|}\geq \dfrac 1 8\\ \dfrac 1 m \geq \dfrac 1 8 \bigcup \dfrac 1 m \leq - \dfrac 1 8\\ m \leq 8 \bigcup m \geq -8\\ m \in [-8,8],~m \in \mathbb{Z}-\{0\}\\ \text{16 total integers}\)

 Please solve the whole problem fully

when this eqaution intercepts the y-axis ....the 'x' value is = 0

put x = 0 in to the equation

y-intercept = 2(0+3)(0-4) = -24

Thank you!

Last one

6.   2.4^(3x+1) = 9    take the log of  both sides

log (2.4) ^(3x + 1)  = log 9      and we can write

(3x + 1) * log 2.4  = log 9

3x * log 2.4 + 1 * log 2.4 =  log 9

(3 log 2.4)x = log 9 - log 2.4     divide both sides by 3 log 2.4

x  =  [ log 9 - log 2.4 ] / [ 3* log 2.4 ] ≈  0.5033

Thank you so much, I needed this

4.    4^x  = 10

Take the log of each side

log 4^x  = log 10       and we can write

x * log 4  = 1      divide both sides by log 4

x  =  1 / log 4

5. (1/4)^2x = (1/2)^x

[ (1/2)^2 ]^(2x)  = (1/2)^x

(1/2)^(4x)  = (1/2)^x

4x  = x

4x - x  = 0

3x  = 0

x  = 0

Are you trying to solve for one variable? yes sorry

1. 4^2x = 32^1/2  ......the idea is to get the bases the same and then solve for the exponents

(2^2) ^(2x)   = (2^5)^(1/2)

(2)^(4x)  = (2)^(5/2)

4x  = 5/2      divide both sides by 4

x  = 5/8

2. 9^x = 3^x-2

(3^2)^x  =3^(x - 2)

(3)^(2x)  = 3^(x - 2)

2x  = x - 2     subtract x from both sides

x = - 2

3. 2^x = 4^x+1

2^x  = (2^2)^(x + 1)

2^x = (2)^(2x + 2)

x = 2x + 2      subtract 2x from both sides

-1x  = 2         divide both sides by -1

x = -2

\(b=t-a\\ a^2+b^2 = a^2 + (t-a)^2 = \\ 2a^2 - 2at + t^2 = \\ 2\left(a-\dfrac t 2 \right)^2 + \dfrac{t^2}{2}\\ \text{This clearly has a minimum value of }\dfrac{t^2}{2} \text{ at }a = \dfrac t 2\)

OK...

1. Slope at any point ...using implicit differentiation....  2x - 2y - 2xy ' + 8y y '  = 0

y' ( 8y - 2x ] = 2y - 2x

y '  =   [ 2y - 2x ] / [ 8y - 2x ]

2.  At x  = 2....we can find y as

2^2 - 2(2)y  + 4y^2  = 4

4 - 4y + 4y^2  = 84

4y^2 - 4y - 80 = 0

y^2 - y - 20  = 0       factor

(y - 5) (y + 4)  = 0     .....so   y = 5  or y = -4

So....we have the points   ( 2, 5) and (2, -4)

So  the slope at  (2, 5)  =  [2(5) - 2(2) ] / [ 8(5) - 2(2) ]  =   6/36  = 1/6

So...the equation is  y  = (1/6) (x - 2) + 5   

The slope at (2, -4) = [ 2(-4) - 2(2)] / [ 8(-4) -2(2)]  =  -12/ -36  =  1/3

So...the equation is  y  = (1/3)(x - 2) - 4

See the graph here : https://www.desmos.com/calculator/q55m709tum

I don't know the answer to the last one.....sorry  !!!

Thank you!! 

This one is a little tricky, Cupcake   !!!!

Not really kosher, but since we assume that 1 - tan^2x  is not equal to 0, we can multiply both sides by 1 - tan^2x  and get that

2tanx  = 1 - tan^2 x      rearrange as

tan^2 x + 2 tanx  - 1  = 0       let   tan x  =  m     and we have that

m^2 + 2m - 1  = 0

m^2 + 2m  =  1      complete the square on m

m^2 + 2m + 1  = 1 + 1

(m + 1)^2  = 2         take both roots

m + 1  = ±√2

m = ±√2 - 1

So  

tan x  = √2 - 1.....take the arctan (√2 - 1)  = m =   22.5°  =  pi/8 + n*pi

And

tan x  =  - √2 - 1.....take the arctan  (-√2 - 1 )  = x  = -67.5°.....add 180 to this  = 112.5°  = [225/2]°  = (5/4)pi / 2  =

(5/8)pi + n * pi

We are using the fact that   both functions are symmetric about the y axis

So

2                                                          2

 ∫ 12 - 3x^2   dx  is  the same  as    2 * ∫  12 - 3x^2  dx   

-2                                                         0

Another way:

You can split the shape up into a parallelogram and a triangle:

3z + 4iz'  = 1 - 8i

3 (a + bi) + 4i(a - bi)   = 1 - 8i

3a + 3bi + 4ai + 4b  = 1 - 8i

3a+ 4b =  1      ⇒   12a + 16b  = 4

4a + 3b  = - 8 ⇒     -12a - 9b = 24

7b  = 28

b = 4        and   3a + 4(4) = 1   ⇒  3a = - 15  ⇒  a = -5

So

z = -5 + 4i

I think you are needing the quadratic formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a = 1    b = -2    c  = -4       Substitute these values to find the values of 'n' that solve the equation    ( I know ....it says 'x'...but you can put 'n' there)