C1 help Watch

Umairy363
Badges: 0
Rep:
?
#1
Report Thread starter 9 years ago
#1
Hey everyone. Just a few questions.

1 It is given that f(x) = (x - 2)^2 - p(x + 1)

a. Find the values of p for which the equation f(x) = 0 has 2 equal roots

b. Show that, when p = 4, f(x) has a minimum value of -16

c. Given that the curver y = f(x) has a turning point when x = 3, find the value of p


2. Complete the square of y = 10 - 6x - x^2

Sorry i know these sound really easy but i'm stumped.

Thankyou
0
reply
technicolour
Badges: 3
Rep:
?
#2
Report 9 years ago
#2
Which parts are you stuck on? How much of it can you do?
0
reply
Small123
Badges: 13
Rep:
?
#3
Report 9 years ago
#3
(Original post by Umairy363)
Hey everyone. Just a few questions.

1 It is given that f(x) = (x - 2)^2 - p(x + 1)

a. Find the values of p for which the equation f(x) = 0 has 2 equal roots

b. Show that, when p = 4, f(x) has a minimum value of -16

c. Given that the curver y = f(x) has a turning point when x = 3, find the value of p


2. Complete the square of y = 10 - 6x - x^2

Sorry i know these sound really easy but i'm stumped.

Thankyou
Starting with a) represent f(x) in quadratic from (ax^2+bx+c) then equate values of p for which the discriminant >0.
0
reply
The Muon
Badges: 2
Rep:
?
#4
Report 9 years ago
#4
(Original post by Umairy363)
Hey everyone. Just a few questions.

1 It is given that f(x) = (x - 2)^2 - p(x + 1)

a. Find the values of p for which the equation f(x) = 0 has 2 equal roots

b. Show that, when p = 4, f(x) has a minimum value of -16

c. Given that the curver y = f(x) has a turning point when x = 3, find the value of p


2. Complete the square of y = 10 - 6x - x^2

Sorry i know these sound really easy but i'm stumped.

Thankyou
Small has done 1 and I cant do any better but here is 2.

to complete the square you need it in the form a(x+b)^2 + c

So just multiply out the brackets and compare coefficients. For example you would have

a(x+b)^2 + c  = 10 - 6x - x^2

ax^2 = -x^2

a=-1

Do the same for the x's and the constants and your onto a winner.
0
reply
Small123
Badges: 13
Rep:
?
#5
Report 9 years ago
#5
(Original post by Umairy363)
Hey everyone. Just a few questions.

1 It is given that f(x) = (x - 2)^2 - p(x + 1)

a. Find the values of p for which the equation f(x) = 0 has 2 equal roots

b. Show that, when p = 4, f(x) has a minimum value of -16

c. Given that the curver y = f(x) has a turning point when x = 3, find the value of p


2. Complete the square of y = 10 - 6x - x^2

Sorry i know these sound really easy but i'm stumped.

Thankyou
b) complete the square so find the minimum value.

c) again complete the square and use the value of p to determine turning point.
0
reply
Umairy363
Badges: 0
Rep:
?
#6
Report Thread starter 9 years ago
#6
(Original post by Small123)
b) complete the square so find the minimum value.

c) again complete the square and use the value of p to determine turning point.
I managed to get 'b' but not 'c'. I don't have p, i hae x, but i've tried putting it in f(x) and i'm getting 17/4, the actual answer is 2 ??
0
reply
Small123
Badges: 13
Rep:
?
#7
Report 9 years ago
#7
(Original post by Umairy363)
I managed to get 'b' but not 'c'. I don't have p, i hae x, but i've tried putting it in f(x) and i'm getting 17/4, the actual answer is 2 ??
Are you allowed to use calculus?
0
reply
Umairy363
Badges: 0
Rep:
?
#8
Report Thread starter 9 years ago
#8
(Original post by Small123)
Are you allowed to use calculus?
I would have used calculus, if i was allowed, i'm a review exercise from the beginning of the book, and they havent taught us calculus yet, in the book so i'm assuming no.
0
reply
Small123
Badges: 13
Rep:
?
#9
Report 9 years ago
#9
(Original post by Umairy363)
I would have used calculus, if i was allowed, i'm a review exercise from the beginning of the book, and they havent taught us calculus yet, in the book so i'm assuming no.
Interesting. Ill give it a try and report back
0
reply
Small123
Badges: 13
Rep:
?
#10
Report 9 years ago
#10
(Original post by Umairy363)
I would have used calculus, if i was allowed, i'm a review exercise from the beginning of the book, and they havent taught us calculus yet, in the book so i'm assuming no.
Alright could you post your working because I get 2.
0
reply
Umairy363
Badges: 0
Rep:
?
#11
Report Thread starter 9 years ago
#11
(Original post by Small123)
Alright could you post your working because I get 2.

f(x) = x^2 + 4 - 4x - p(x+1)
-16 = 9+4 - 12 - px - p
-17 = -3p-p
-17 = -4p
p = 17/4
0
reply
Small123
Badges: 13
Rep:
?
#12
Report 9 years ago
#12
(Original post by Umairy363)
f(x) = x^2 + 4 - 4x - p(x+1)
-16 = 9+4 - 12 - px - p
-17 = -3p-p
-17 = -4p
p = 17/4
You dont use part b to help you for this one.
Rearranging f(x) to give you x^2 - (p+4)x -p+4. Then complete the square and find the value of p which yields a min at x=3.
0
reply
Umairy363
Badges: 0
Rep:
?
#13
Report Thread starter 9 years ago
#13
(Original post by Small123)
You dont use part b to help you for this one.
Rearranging f(x) to give you x^2 - (p+4)x -p+4. Then complete the square and find the value of p which yields a min at x=3.
oh. Ok i see, thankyou very much.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (302)
37.47%
No - but I will (61)
7.57%
No - I don't want to (60)
7.44%
No - I can't vote (<18, not in UK, etc) (383)
47.52%

Watched Threads

View All
Latest
My Feed