Question

9.Given the minimum value of the quadratic function f(x)=2x^2+px+q is 1 when x=-3,find the value of p and of q. @ganeshie8 @iambatman @Jhannybean @mathmath333

Answer 1

i got p=12 and q=37 not sure whether my answer is right

Answer 2

\[\frac{ dy }{ dx } = 4x + p\]
and the minimum value is when derivative is zero so we can say \( 4x + p = 0 \) and we know that \(x=-3 \) so \( p = 12 \)

Answer 3

\[\frac{ dy }{ dx }~?\]

Answer 4

it's derivative of that function,if u have not studied it...remember this point,the equation \( ax^2 + bx + c \) when \(a >0\) has a minimum point \[(\frac{ -b }{ 2a } , \frac{ 4ac - b^2 }{ 4a })\]

Answer 5

here \(\large \frac{ -b }{ 4 } = -3\) so \( b = 12 \)

Answer 6

can we use complete square method? @PFEH.1999

Answer 7

this way is easier and the point i said u comes from square method but if u want to use it it would get confusing...

Answer 8

bcoz i haven't learn \[(\frac{ -b }{ 2a },\frac{ 4ac-b^2 }{ 4a })\]

Answer 9

well then,use square method...you'll reach the same thing.

Answer 10

okay

Answer 11

i got p=12 and q=37 but i'm not sure whether i'm right. @PFEH.1999

Answer 12

let me check...

Answer 13

Okay

Answer 14

would u check ur work again?

Answer 15

i got p=12 and q=37 @PFEH.1999

Answer 16

ok write ur solution i got q=19

Answer 17

\[f(x)=(2x+\frac{ p }{ 2 })^2-\frac{ p^2 }{ 4 }+q\]
\[=2(x+\frac{ p }{ 4 })^2-\frac{ p^2 }{ 4 }+q\]
\[=2(x+3)^2-36+q\]
\[=2(x+3)^2+1\]
\[\frac{ p }{ 4 }=3\]
\[p=12\]
\[-36+q=1\]
\[q=37\]

Answer 18

@PFEH.1999

Answer 19

http://www.wolframalpha.com/input/?i=1%3D2x%5E2%2Bpx%2Bq%2Cx%3D-3

Answer 20

put \(p=12\) and find \(q\)

Answer 21

i got q=19 @mathmath333

Answer 22

me too got \(19\)

check it with your book answer

Answer 23

the book answer written q=19 @mathmath333

Answer 24

\(\Huge \bf done!\)

Answer 25

can u show me the full working @mathmath333 ?

Answer 26

\(\large\tt \begin{align} \color{black}{
f(x)=2x^2+px+q\\~\\
\text{f(x)=1,when x=-3, so}\\~\\
1=2(-3)^2+p(-3)+q\\~\\
3p-q=17-----\color{red}{(1)}\\~\\
f(x)=2x^2+px+q\\~\\
y=2x^2+px+q\\~\\
\frac{ dy }{ dx } = 4x + p\\~\\
0 = 4x + p\\~\\
0 = 4(-3) + p\\~\\
p=12\\~\\
\text{put p=12 in (1) and get "q"}
}\end{align}\)

Answer 27

\(\bf\huge\color{#ff0000}{T}\color{#ff2000}{h}\color{#ff4000}{a}\color{#ff5f00}{n}\color{#ff7f00}{k}~\color{#ffaa00}{Y}\color{#ffd400}{o}\color{#bfff00}{u}\color{#4600ff}{!}\color{#6800ff}{!}\color{#8b00ff}{!}\) @mathmath333

Answer 28

and @PFEH.1999

Answer 29

@MARC_ sry alot ;) my internet connection suddenly broke up and i couldn't answer any more.. @mathmath333 thank you for helping him ;)

Answer 30

It's okay. @PFEH.1999 :)