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y = ax^2 + bx +  c

The vertex of the parabola  will  occur  at   [  (-2 + 4) / 2 ,  54 ]  =    [2/2 , 54]  = [ 1, 54]

And the x coordinate of the vertex  =   -b/ [2a ]  = 1     so   b = -2a

So we have this system of equations

a(1)^2  -2a(1) + c =  54    ⇒   -a + c  = 54  ⇒    a - c  = -54

a(-2)^2  -2a(-2) + c  = 0   ⇒  8a + c = 0  ⇒      8a + c   = 0     

Add the last two equations and we get that

9a  =  -54

a  = - 6

b = -2(-6)  = 12

And

8(-6) + c  = 0   ⇒  c = 48

So   a  + b  + c    =    -6 +  12  +  48  =    54

Here's the graph :  https://www.desmos.com/calculator/9u0epc5dbu

Dang i didn't know i have this many fans jeez no need to trip everyone can get a autograph 

Thanks :)

Sorry for all these questions XD Plz forgive me XD. But thanks :)

Thanks.....there is still a LOT more I have to learn though....

Very nice, hectictar......you are very good at this sort of thing  !!!

First one

h(7)  means that we go to  "7"  on the  x axis....from here....we go up until we intercept the graph and we read the y value at that point....so.....going out to 7  and then going up we find that we intercept the graph at y = 5

So  f(7)  = 5

So  f(f(7))  = f(5)

Then....we do the same thing.....go to 5 on the x axs  and then go down until we intercept the graph at the y value of  -1

So...   f (f(7)) =  f(5)  =  -1

Second one

f^-1 (4)   means that we  go up  on the graph until y = 4  and then we find the x value associated with that y value

So...when we go up to y = 4.....the x value asociated with this y value  is 3

So

f^-1(4)  =   3

I am confused. What?

\(\dfrac{z\overline z}{w\overline w}\)

So we have

[-8 + 15 i ] [ -8 - 15i ]               64  - 225i^2            64 - 225(-1)           289

_________________  =       ___________  =    ___________  =    ______

[ 6 - 8i ] [ 6 + 8i ]                     36 - 64i^2              36  - 64(-1)              100

Sorry....I mis-copied (1)....see my edit

However...(3)  is correct....see the graph, here : https://www.desmos.com/calculator/zvlbv8gf1t

When we have  k  distinguishable  balls   and  n  distinguishable boxes....and assuming that either box might be empty....then the number of ways to distribute the balls is given by :

n^k  =  (2)⁵  =    32 ways

When you think about it....if we looked at only Box1, each consignment of balls in this box would just be a subset of the number of total subsets formed by  5 objects in a set  = 2⁵

I'm sorry, all the answers are incorrect except for 4a and 4b.

There aren't any questions or specifics here.  So, there is nothing to answer.

11 apples - 1 apple  =  10 apples

11 thirds - 1 third  =  10 thirds

\(\dfrac{11}{3}\ -\ \dfrac13\ =\ \dfrac{10}{3}\)

No, this is a math question that I have to ask.

1. Consider the parabola y = -5x^2+17x-12 . What is the vertex of this parabola?

We have the form Ax^2 + Bx + C

The x coordinate of the vertex  is given by   -B / [2A]  =  -17/ [ 2(-5)]  =  17/10

Put this value back into the function to get the y coordinate of the vertex 

-5(17/10)^2 + 17(17/10) - 12  =

-5 (289/100) + 289/10 - 12   =   49/20

So.....the vertex  is  ( 17/10, 49/20)

2. What is the equation of the line of symmetry of the parabola y=-5x^2+17x-12

Easy.....the  equation for the symmetry line is 

x  = x coordinate of the vertex

So....   x  = 17/10

3.What are the x-coordinate(s) of all point(s) where the parabola y=-5x^2+17x-12 intersects the line y=0?

We need to solve this

-5x^2 + 17x - 12  =  0     factor as

(-5x + 12) ( x - 1)  = 0

Set each factor to 0  and solve for x   and we have that

-5x + 12 = 0            x  - 1  = 0

12 = 5x                    x = 1

12/5 = x        

4. For what values of x  is -5x^2+17x-12 negative?

The parabola has its vertex above the x axis  and it turns downward because of the -5 in front of the x^2 term

It will intersect the x axis at  x = 1  and x  = 12/5

So......it will be negative from  (-infinity , 1)  and from (12/5, infinity)

See the graph here : https://www.desmos.com/calculator/visfr9nhfz