Question

Part 1: Write the general form of the equation which matches the graph below.

Part 2: In complete sentences, explain the process taken to find this equation
PLease help me! Im confused!

Answer 1

this is a direct correlation to the other one we did

Answer 2

how?

Answer 3

notice it gives you an "a" measure, and a definable center point

first thing to do is to write the general setup;

(y-k) = -4a (x-h)^2

Answer 4

so far I think Im doing it right. I have (y-0)^2=4a(x-3)^2

Answer 5

okay I did that. Is that right?

Answer 6

that great except that this parabola opens down so we want a negative on it

Answer 7

if we define the "a" with a direction of down we can assume 4(-a) if we want ....

Answer 8

how far is A from the center?

Answer 9

(3,-4) away from (0,0) is that what I need

Answer 10

?*

Answer 11

(0,0) is the origin point of the graphing plane, the center of the parabola is (3,0)

how far is (3,-4) from (3,0) ?

Answer 12

4 away. its 4 down

Answer 13

good, then |a| = 4, so lets define a with direction to be -4

(y-0)=4a(x-3)^2
-4
---------------
y = -16(x-3)^2

Answer 14

i believe the general form would have us expand this out into:

y = ax^2 + bx + c

Answer 15

so would you distribute the -16 then?

Answer 16

we cant do that just yet ... or rather, there is an easier way to step thru this

Answer 17

ohh

Answer 18

(x-3)^2 = (x-3) (x-3)

can you give me the expanded form of that?

Answer 19

x^2-6x+9

Answer 20

very good; now we can distribute the -16 thru wihtout having to worry about that ^2 getting in the way

Answer 21

-16(x-3)^2 = -16 (x^2-6x+9)

Answer 22

ohh okay. then it becomes -16x^2 + 96x-144 ?

Answer 23

...... ugh, i think i had my wires crossed again :/

the mathing is right according to what we have done, but our original setup is flawed

Answer 24

4ay = x^2 ..... not y=4a x^2

Answer 25

so how does that change everything?

Answer 26

when you start out wring, you end up wrong :/ but since we covered the steps to get here, lets do a quick run thru of the proper setup

Answer 27

okay

Answer 28

a = -4, (3,0) -> y, x-3

4(-4)y = (x-3)^2

-16 y = x^2 -6x +9

y = -(x^2 -6x +9)/16

Answer 29

try to think of a way to remember that 4a is not ^2 :) that way you can correct me the next time lol

Answer 30

haha okay so y=-(x^2-6x+9)/16 is the answer?!

Answer 31

that is the results, im not sure if they want that into 3 separate terms or if that is sufficient. Depends on the grader

Answer 32

I think thats fine! thankyou sooo much! I think I get itnow!

Answer 33

youre welcome ;)