Question

What is the quadratic function that is created with roots -10 and -6 and a vertex at (-8, -8)

Answer 1

subtract each root by x, and provide a scalar "s" to adjust the results with

y = s (-10-x) (-6-x) ; such that when x=-8, y=-8

Answer 2

what?

Answer 3

.... see post above :/

Answer 4

i saw that post above

Answer 5

then youll have to be more concise as to what you are confused about

Answer 6

subtract each root by x

(-10-x) (-6-x) ; what does this expand into?

Answer 7

-10x and -6x right

Answer 8

i think i have no idea how to do this at all

Answer 9

postives, but yes

x^2 + 16x +60

Answer 10

now, when x=-8, we need to know what this produces, so taht we can know what to scale it by to get -8 back out

Answer 11

ok

Answer 12

x^2 + 16x +60; when x=-8 would be?

Answer 13

might be easier to work out in the factored form

(-10--8) (-6--8)

(-2) (2) = -4

Answer 14

-8^2 + 16(8) +60
and i have no clue what you just did there

Answer 15

is what i did correct without the simplifying part

Answer 16

i inserted -8 into the factored form i started with, just makes life a little simpler :)

Answer 17

yes, the math will work it the same

Answer 18

so it is -64+(-128)+60 is that correct

Answer 19

almost, remember to include all of x into the squared part like this

(-8)^2 = 64
+16(-8) = -128
+60 = + 60

Answer 20

so you add the 60 to the -128 correct

Answer 21

yes, then add the 64 to finish it off

Answer 22

so that would be -4

Answer 23

correct

Answer 24

good; now, we want a scalar, a multiplier, such that -4 times ? = -8

-4s = -8; what is s?

Answer 25

wait i got another question it says to Enter your answer as a function beginning with f(x).
and -4 correcct

Answer 26

*correct

Answer 27

not yet, lets determine the equation first, then we will worry about syntax

Answer 28

ok

Answer 29

so the answer to your question is -4 right?

Answer 30

as is: x^2 +16x + 60 gives us -4 when x=-8, but we dont want a -4 ... we want a -8; so the only thing we can do at this point is find a scalar to modify our equation with such that

-4s = -8

this will give us the equation that we want

sx^2 +16sx +60s

so what is "s"?

Answer 31

wait how do you find out s though

Answer 32

by solving this: -4s = -8

-4 times "what" is equal to -8

Answer 33

2

Answer 34

can you tell me why we need to do that?

Answer 35

no

Answer 36

here is why: we are asked

What is the quadratic function that is created with roots -10 and -6 and a vertex at (-8, -8)

we found the root parts: x^2 + 16x + 60 satisfies the roots; but it does not satisfy the vertex

when x=-8, we want y=-8; at the moment we got x=-8, y=-4; so we need to find a scalar to multiply out setup by to get us to the required y=-8


s = 2 is what we need since -4*2 = -8

therefore our function is: 2x^2 +32x +120

Answer 37

what how did you get the function thing

Answer 38

is that the answer right there? if so this is the example answer Format using this sample answer:f(x)=(3)(x^2-21x-19) of how it is supposed to be so would it be 2(x^2+32x+120)?

Answer 39

... we just worked it from start to finish :/

subtract each root by x, and provide a scalar "s" to adjust the results with

y = s (-10-x) (-6-x) , s=2

y = 2 (-10-x) (-6-x)
= 2(x^2+16x+60)
= 2x^2+32x+120

Answer 40

ohhh ok that makes since so what about my last question

Answer 41

well, we have 3 equivalent setups to choose from, and it looks like the middle setup matches your required syntax

Answer 42

f(x) = 2 (x^2+16x+60)

Answer 43

ok thank you

Answer 44

good luck ;)