Ok i'm a bit stuck at this question, so help would be appreciated

a) Show that the function f(x) = 2x^2 + 8x -3 can be written in the form

f(x) = a (x+b)^2 + c where a,b, and c are constants.

b) Hence, or otherwise, find the coordinates of the turning point of the function f.

**0**

# Function

Started by Allana, Sep 04 2005 04:00 PM

7 replies to this topic

### #1

Posted 04 September 2005 - 04:00 PM

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### #2

Posted 04 September 2005 - 05:15 PM

ok this is exactly the same idea as the ones Rocky put up before

You use the method of "Completing the Square" which you probably wont know but that is not a problem because the questions is worded so you dont

anyway you start with f(x) = 2x

basically want i am doign here is looking at the form i wand seeing what i can do to get there

for the a well it would be easy to take the 2 out from the terms in X

2(x

next is to get the bracket squared part

for this we need something in the form ax

the c part is found from the formula (.5 * b) squared which gives 16

so we now have

2(x

2(x

2(x

tidy it up we get

2(x+4)

now for the next bit

a turning point will be where the function is at a max or min value

if u look at the function

if we remove all the terms in x we are left with -19

this is our min

so what value of x gives us a y = -19

answer is x = 4

You use the method of "Completing the Square" which you probably wont know but that is not a problem because the questions is worded so you dont

anyway you start with f(x) = 2x

^{2}+8x -3basically want i am doign here is looking at the form i wand seeing what i can do to get there

for the a well it would be easy to take the 2 out from the terms in X

2(x

^{2}+ 4x) - 3next is to get the bracket squared part

for this we need something in the form ax

^{2}bx + cthe c part is found from the formula (.5 * b) squared which gives 16

so we now have

2(x

^{2}+ 8x + 16) -3 , however now we have just added a 16 from nowhere so to even things up we take the 16 away again2(x

^{2}+ 8x + 16) -3 -162(x

^{2}+ 8x + 16) -19tidy it up we get

2(x+4)

^{2}- 19now for the next bit

a turning point will be where the function is at a max or min value

if u look at the function

if we remove all the terms in x we are left with -19

this is our min

so what value of x gives us a y = -19

answer is x = 4

If i am not here i am somewhere else

### #3

Posted 04 September 2005 - 05:47 PM

QUOTE(Allana @ Sep 4 2005, 05:00 PM)

Ok i'm a bit stuck at this question, so help would be appreciated

a) Show that the function f(x) = 2x^2 + 8x -3 can be written in the form

f(x) = a (x+b)^2 + c where a,b, and c are constants.

b) Hence, or otherwise, find the coordinates of the turning point of the function f.

a) Show that the function f(x) = 2x^2 + 8x -3 can be written in the form

f(x) = a (x+b)^2 + c where a,b, and c are constants.

b) Hence, or otherwise, find the coordinates of the turning point of the function f.

The equation can be written in the other form by 'completing the square'.

Here's the working (I get a different answer from Dave though ):

= (seperating the constant c)

= (taking out 2 as a common factor)

=

=

=

The t.p is

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### #4

Posted 04 September 2005 - 05:53 PM

hmmm I still don't really understand the first part.

### #5

Posted 04 September 2005 - 06:58 PM

i didnt take the 2 out of the 8x when i did it on paper

If i am not here i am somewhere else

### #6

Posted 04 September 2005 - 07:15 PM

QUOTE(Allana @ Sep 4 2005, 06:53 PM)

That'd be because ya probs haven't done stuff on completing the square yet. Once you get to quadratics, you'll understand.

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### #7

Posted 04 September 2005 - 07:21 PM

QUOTE(Wedge37 @ Sep 4 2005, 07:15 PM)

QUOTE(Allana @ Sep 4 2005, 06:53 PM)

That'd be because ya probs haven't done stuff on completing the square yet. Once you get to quadratics, you'll understand.

Ohhh. Well I guess it has to be handed in for Thursday so that makes sense

### #8

Posted 04 September 2005 - 10:09 PM

Page 6 of the Unit 2 Outcome 1download has notes for 'completing the square'

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