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8x-x^2 can be written in the form p-(x-q)^2, for all values of x:

a) Find value of p and q

b) Expression 8x-x^2 has a maximum value.

bi) Find maximum value

bii) State value of x for which the maximum value occurs

Also how can u solve the simultaneous equations:

4x^2-y^2=40

2x+y=10

a) Find value of p and q

b) Expression 8x-x^2 has a maximum value.

bi) Find maximum value

bii) State value of x for which the maximum value occurs

Also how can u solve the simultaneous equations:

4x^2-y^2=40

2x+y=10

a) Expand $p-{(x-q)}^2$ and you should be able to see first what q is, and then using that to help you work out what p is

b) i) Using your factorised version, what must be in the brackets to make $p-{(x-q)}^2$ as big as possible? Use that to work out the maximum value

ii) You know numerically what's in the brackets, and also in terms of x and a constant (q)

For your simultaneous equations, rearrange the second to give you either x in terms of y or y in terms of x (I suggest y in terms of x, there's no division). Substitute that y into the first equation to get out a quadratic in x. Solve this to get two values of x. Substitute those two values of x into the second simultaneous equation to get y - do one x at a time and pair it with the y that you get out, usually in the form $(x_1\,y_1)$, $(x_2\,y_2)$

b) i) Using your factorised version, what must be in the brackets to make $p-{(x-q)}^2$ as big as possible? Use that to work out the maximum value

ii) You know numerically what's in the brackets, and also in terms of x and a constant (q)

For your simultaneous equations, rearrange the second to give you either x in terms of y or y in terms of x (I suggest y in terms of x, there's no division). Substitute that y into the first equation to get out a quadratic in x. Solve this to get two values of x. Substitute those two values of x into the second simultaneous equation to get y - do one x at a time and pair it with the y that you get out, usually in the form $(x_1\,y_1)$, $(x_2\,y_2)$

rearrange 2x+y=10 into y= ... and then substitute into the other equation. When you find x, sub the value back into 2x+y=10 to get your answer.

ye but i dont know how to compare it can sum1 PLEASE type out all the working out

thnx 4 all ur replies i understand it now

Original post by *Princess*

thnx 4 all ur replies i understand it now

i have this same question and i know how to do part a) but i do not understand part b) i) find the maximum value of 8x-x^2 and ii) state the value of x for which this maximum occurs.

can you please help me

This is quite old thread!

I was going to quote the posters to see if my answer is correct, but I doubt they'd come on anymore!

I think this is what you do:

$16-(x-4)^2$

The maximum value of $8x-x^2$ is -16.

The maximum value of $x$ is 4.

The general formula for completing squares is:

$a(x+h)^2+k$

$a$ is the maximum, but this needs to be a negative number. If the $a$ value in the equation if negative, leave it as it is, if it is positive, change it to negative.

$h$ is your maximum value for $x$.

$k$ is the maximum value for $y$.

So in this equation, the maximum coordinates are $(4, 0)$ because there is no $k$ value therefore it must be 0. It is 4 not -4 because you're meant to do the opposite of what is in the brackets, like how $y=f(x+2)$ you move to the left rather than to the right and how $y=f(2x)$ is actually a stretch by scale factor $\frac {1}{2}$.

So the answers are 16 and 4.

I'm not 100% sure on this though...

(This is based upon this post on yahoo answers.)

Other links:

Wikipedia

Youtube

Edit: Now I've just confused myself about maximum and minimum - although I've found the answers and this is correct, however it seems that many sources online say this is how you work out a minimum? :/

I was going to quote the posters to see if my answer is correct, but I doubt they'd come on anymore!

I think this is what you do:

$16-(x-4)^2$

The maximum value of $8x-x^2$ is -16.

The maximum value of $x$ is 4.

The general formula for completing squares is:

$a(x+h)^2+k$

$a$ is the maximum, but this needs to be a negative number. If the $a$ value in the equation if negative, leave it as it is, if it is positive, change it to negative.

$h$ is your maximum value for $x$.

$k$ is the maximum value for $y$.

So in this equation, the maximum coordinates are $(4, 0)$ because there is no $k$ value therefore it must be 0. It is 4 not -4 because you're meant to do the opposite of what is in the brackets, like how $y=f(x+2)$ you move to the left rather than to the right and how $y=f(2x)$ is actually a stretch by scale factor $\frac {1}{2}$.

So the answers are 16 and 4.

I'm not 100% sure on this though...

(This is based upon this post on yahoo answers.)

Other links:

Wikipedia

Youtube

Edit: Now I've just confused myself about maximum and minimum - although I've found the answers and this is correct, however it seems that many sources online say this is how you work out a minimum? :/

(edited 13 years ago)

Original post by drewb

This is quite old thread!

I was going to quote the posters to see if my answer is correct, but I doubt they'd come on anymore!

I think this is what you do:

$16-(x-4)^2$

The maximum value of $8x-x^2$ is -16.

The maximum value of $x$ is 4.

The general formula for completing squares is:

$a(x+h)^2+k$

$a$ is the maximum, but this needs to be a negative number. If the $a$ value in the equation if negative, leave it as it is, if it is positive, change it to negative.

$h$ is your maximum value for $x$.

$k$ is the maximum value for $y$.

So in this equation, the maximum coordinates are $(4, 0)$ because there is no $k$ value therefore it must be 0. It is 4 not -4 because you're meant to do the opposite of what is in the brackets, like how $y=f(x+2)$ you move to the left rather than to the right and how $y=f(2x)$ is actually a stretch by scale factor $\frac {1}{2}$.

So the answers are 16 and 4.

I'm not 100% sure on this though...

(This is based upon this post on yahoo answers.)

Other links:

Wikipedia

Youtube

Edit: Now I've just confused myself about maximum and minimum - although I've found the answers and this is correct, however it seems that many sources online say this is how you work out a minimum? :/

I was going to quote the posters to see if my answer is correct, but I doubt they'd come on anymore!

I think this is what you do:

$16-(x-4)^2$

The maximum value of $8x-x^2$ is -16.

The maximum value of $x$ is 4.

The general formula for completing squares is:

$a(x+h)^2+k$

$a$ is the maximum, but this needs to be a negative number. If the $a$ value in the equation if negative, leave it as it is, if it is positive, change it to negative.

$h$ is your maximum value for $x$.

$k$ is the maximum value for $y$.

So in this equation, the maximum coordinates are $(4, 0)$ because there is no $k$ value therefore it must be 0. It is 4 not -4 because you're meant to do the opposite of what is in the brackets, like how $y=f(x+2)$ you move to the left rather than to the right and how $y=f(2x)$ is actually a stretch by scale factor $\frac {1}{2}$.

So the answers are 16 and 4.

I'm not 100% sure on this though...

(This is based upon this post on yahoo answers.)

Other links:

Wikipedia

Youtube

Edit: Now I've just confused myself about maximum and minimum - although I've found the answers and this is correct, however it seems that many sources online say this is how you work out a minimum? :/

Thank-you so much. I actually get it eventhough i thought it was very confusing at first. I also checked out the links that you attached and they were really helpful.

I tried the steps for a similar question like this and got it right

thanks again

- Maths help
- A Level Maths Exam Question
- Quadratics
- Converting scale to feet
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- circle question help
- Mathswatch
- Bmo/smc ukmt
- i rotted my potential away
- mathematics
- The Pupillage Interview/Acceptance/Rejection Thread 2024 Watch
- Need help on log question urgent, mock exam tomorrow
- acids and bases question
- Bmo 2023/24
- AQA Level 2 Further Maths 2024 Paper 1 (8365/1) - 11th June [Exam Chat]
- Wjec further maths unit 1 thoughts
- ah maths help
- Circles- maths
- iDEA Silver Maker Badge Activation
- Capacitors

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