Question

F(x)=9-8x-x^2..?

Answer 1

What you want to do with F(x)??

Answer 2

How do you find the vertex, its domain and range, and x-intercepts?

Answer 3

See, x intercept you can find by putting F(x) = 0 and then find the value of x.

I don't know about the others..

Others will definitely help you in finding them..

Answer 4

can you identify a, b, and c when you write the function in this form:
\[\huge f(x)=ax^2+bx+c \]

Answer 5

when you do, the x-coordinate of the vertex is obtained by: \[\huge x=-\frac{b}{2a} \]

Answer 6

to get your y-coordinate of the vertex, just plug that value in to f.., that is, find the value of
\[\huge f(\frac{-b}{2a}) \]

Answer 7

Is x 4/9? @dpaInc

Answer 8

nope..
a =
b =
c =

Answer 9

a is -1 and b is -8, and c is 9?

Answer 10

4?

Answer 11

yes...
so,
\[\huge x=-\frac{b}{2a}=-\frac{(-8)}{2(-1)}=...\]

Answer 12

yes.... x=4..

Answer 13

what's the y-value?

Answer 14

How do you know what the y-value is? Sorry, I'm not good with this stuff

Answer 15

y = f(4)

Answer 16

y = f(4) = \(\large 9-8(4)-(4)^2 \)=

Answer 17

-39?

Answer 18

so the point (4, -39) is your vertex...

Answer 19

now which way does this parabola open?

Answer 20

that's determined by the "a" coefficient... if a>0, opens up... if a<0, opens down.

Answer 21

So do you need to use the vertex formula?

Answer 22

no... you can just use the "a" from what I wrote above....

Answer 23

is "a" positive or negative?

Answer 24

What is a? Is it 4 or -39??

Answer 25

a is the coefficient of the x^2 term...

Answer 26

16?

Answer 27

remember when i asked
a =
b =
c =
??

what did you have for a?

Answer 28

1

Answer 29

a is positive.

Answer 30

look again... you told me a=-1....

Answer 31

I'm sorry, yes, it's a negative.

Answer 32

since a<0, the parabola opens downward....