Question

Consider the following function.

Answer 1

\[p(x) = -x ^{2} + 8x - 12\]

Find two points on the graph of the parabola other than the vertex and x-intercepts.

A: ( , )

B: ( , )

Answer 2

@vocaloid

Answer 3

Just plug in any number for x into the equation and simplify it. That'll be the y-value of that point.

Answer 4

could you show me ?

Answer 5

For example, if you plug in x = 1

p(1) = -(1)^2 + 8(1) - 12

1^2 is 1
8 * 1 = 8

p(1) = -1 + 8 - 12
p(1) = -19

so one of the points would be (1, -19)

Answer 6

the x-value we chose was 1
and the y-value you get when you plug it in is -19

thus, (1, -19)

Answer 7

ohh okay , so i use -1 ?

Answer 8

p(-1) = -(1)^2 + 8(-1) - 12

is this right ?

Answer 9

P = 21

Answer 10

where'd your negative sign go?

Answer 11

I thought -1 would be the next one to use for the second potential point

Answer 12

p(-1) = -(1)^2 + 8(-1) - 12
p(-1) = -1 - 8 - 12
p(-1) = -21

(-1,-21)

you dropped the negative signs

Answer 13

okay , is the first point correct as it is ?

Answer 14

yes

Answer 15

okay thank you

Answer 16

it said incorrect for some reason

Answer 17

oh man , it's alright. i have another one

Answer 18

OH MY

Answer 19

I CAN'T ADD OR SUBTRACT

Answer 20

I DONE GOOFED

Answer 21

I HAVE BECOME THE VERY THING I SWORE TO DESTROY

Answer 22

😶

Answer 23

Consider the following function.

\[a(x) = -x^{2} + 8x - 15\]

Find two points on the graph of the parabola other than the vertex and x-intercepts.

Answer 24

it's cool

Answer 25

do the same thing; pick a x coordinate and solve for y

Answer 26

okay i'll need you to check it if it's right

Answer 27

let's try 2

a(2) = (2 - 8)^2 - 15

Answer 28

are you sure

-x^2 + 8x - 15
plugging x as 2 would be
\(-(2^2) + 8(2) - 15\)

Answer 29

like that ?

Answer 30

well, simplify it now

Answer 31

-3

Answer 32

yeah so what's the coordinate point?

Answer 33

2, -3

Answer 34

yes, now use another x value

Answer 35

3

Answer 36

\[-(3^2) + 8 (3) - 15 \]

correct ?

Answer 37

yeah

Answer 38

actually you can't use that point because it's a x intercept..
both (3,0) and (5,0) are x intercepts

so choose another x value

Answer 39

Ok , 4

Answer 40

how about 4 ?

Answer 41

find y

Answer 42

I have -3, 0, and 1

Answer 43

?

Answer 44

for y

Answer 45

-3 to use as y ?

Answer 46

at x=4... what is the value of y?

Answer 47

1

Answer 48

yes, so what's the coordinate point

Answer 49

-8

Answer 50

what....
we have the x and y
(4,1)

Answer 51

i'm sorry, i had -(1^2) + 8(1) - 15

Answer 52

I think you know how to do your questions now

Answer 53

Not really , i just catch on to how you show me the equation to solve it

Answer 54

Was it wrong how i did it ?

Answer 55

no, you're right

Answer 56

for my solution , (1, -8) ?

Answer 57

yeah

Answer 58

ok, thank you