Question

Rewrite this quadratic function f(x)=-2x^2 - 20x + 9 in standard form

Answer 1

Multiply the whole quadratic by -1,
Thus,
f(x) = 2x^2 + 20x - 9

Answer 2

and that is standard form?

Answer 3

Yes. The x^2 term should always be positive. The standard form of a quadratic is ax^2 +/- bx +/- c.

Answer 4

That is false. You cannot simply multiply the function by -1 as it becomes a different function than the original. The given function is in standard form. \(a\) is allowed to be negative.

Answer 5

ok i was pulling my hair out trying to see what was going wrong here..........

Answer 6

do you mean to put it in vertex form? 'cause as yakeyglee says, it already is in stadard form...

Answer 7

it is telling me that it is wrong when i put it in as what it is already... it definitely says to put in standard form

Answer 8

then maybe they are referring to vertex-form then. do you wanna begin?

Answer 9

yes
lets begin...

Answer 10

btw, do you know what vertex form is?

Answer 11

i am not sure how to do vertex

Answer 12

i thought vertex was standard to be honest

Answer 13

true... some people call vertex form as their standard form...

Answer 14

you want to get your equation into looking like this:
\[\large f(x) =a(x-h)^2+k\]
where (h,k) is the vertex of your parabola.

Answer 15

Hint: the axis of symmetry of a parabola is at \(-\frac{b}{2a}\), which is found by averaging the values of the zeroes output by the quadratic formula.

Answer 16

so = a(x-20)+9

Answer 17

f(x) = a(x - 20)^2 + 9 ?

Answer 18

no...
let's use the hint provided by yakeyglee...
can you tell me the x-coordinate of the vertex. use your equation for the values of a and b.

Answer 19

-20/-4

Answer 20

good... so h=5 right?

Answer 21

does that - before the equation cause the negative in the denominator to go away?

Answer 22

the equation -b/2a

Answer 23

yes... so -20/(2*(-2)) = 5

Answer 24

this is the x-coordinate of your vertex.

Answer 25

ok so i can plug h into that equation you gave me.

Answer 26

yes.

Answer 27

so far you have f(x)=-2(x-5)^2 + k

Answer 28

all we gotta do is find k

Answer 29

is k not given with -9?

Answer 30

i mean + 9

Answer 31

no. the k is the y-coordinate of the vertex.
so find f(5) =

Answer 32

-10 - 5^2 + k

Answer 33

hang on man..... i was working with the wrong equation...

Answer 34

i was working with -2x^2+20x-9 when it should have been -2x^2-20x-9

Answer 35

why does that sign change from 9 to -9

Answer 36

i got it wrong again.. sorry.... let's be clear... is this the function:
f(x) = -2x^2 - 20x + 9

Answer 37

that is the function. yes

Answer 38

-b/2a = -(-20)/(2*(-2)) = -5
this is h. so h=-5 correct?

Answer 39

-5 ok thats what i was wondering before.. yes.

Answer 40

ok... we're good now... all we have to do is find k.

Answer 41

to find k, you need to find f(h) = f(-5) =

Answer 42

so F(h) = a(x+5)^2+k .

Answer 43

what's k? find f(-5)

Answer 44

plug in x=-5 into your equation...

Answer 45

do you need help?

Answer 46

ok i dont know what i did wrong.. i just got = a^2+k .. maybe im just not the best at evaluating

Answer 47

f(-5) = -2*(-5)^2 - 20*(-5) + 9

Answer 48

59

Answer 49

ok... that's your k..
so
a = -2
h = -5
k = 59

\[\large f(x)=-2(x-(-5))^2+59=-2(x+5)^2+59\]

Answer 50

try that...

Answer 51

that was it ! .. wow..

Answer 52

so that equation -b/2a

Answer 53

yes... that's to find the x coordinate of your vertex

Answer 54

you can whatever that is to f and solve to get k... then solve from there

Answer 55

set*

Answer 56

yes.

Answer 57

that is very helpful.. i can probably do these on my own now. thanks a lot

Answer 58

sorry 'bout my messing up... could've finished this way faster if it wasn't for me... thanks for not getting pissed...

Answer 59

yw... :)

Answer 60

thank you yakey too, and yeah i just really wanted to learn how to do it, frustrated or not