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1. how do i find the minimum/maximum point of this equation:

x2-8x+15

the 2 being the squared
2. (Original post by maccaism)
how do i find the minimum/maximum point of this equation:

x2-8x+15

the 2 being the squared
What do you think you would need to do to the equation?

(Think about previous questions or examples you've seen).
3. -3 -5

so x=+3 and +5

Not sure tbh by what you mean.
4. (Original post by maccaism)
how do i find the minimum/maximum point of this equation:x2-8x+15the 2 being the squared
Differentiate the original equation (dy/dx) , then equate the new equation to 0 and solve.

EDIT: to find the x values
5. (Original post by maccaism)
how do i find the minimum/maximum point of this equation:

x2-8x+15

the 2 being the squared
Complete the square is probably what you want.

Differentiating works but I don't think that's the aim of the Q
6. you could find the roots... the turning point is always equidistant from the roots on a quadratic function
7. (Original post by 08Mercyf)
-3 -5

so x=+3 and +5

Not sure tbh by what you mean.
i got that............. so is that the minimum point?
8. isn't the minimum point completing the square so (x+ 4)^2 + 15 which becomes (x+4)^2 -1 , therefore the Min value is -1 ??? No point differentiating it. This type of question comes up in a lot of past papers and this is usually what they ask you to do when finding the min/max value.
9. There is only one point as it is a quadratic equation.
10. (Original post by ihatehannah)
isen't the minimum point completing the square so (x+ 4)^2 + 15 which becomes (x+4)^2 -1 , therefore the min value is -1 ???
yep i did completing the square and i got that answer too. i just forgot how to find the minimum value and the x and y co ordinates.

so the minimum value is -1?
and would the x and y co ordinates be 3,5?
11. (Original post by ihatehannah)
isen't the minimum point completing the square so (x+ 4)^2 + 15 which becomes (x+4)^2 -1 , therefore the min value is -1 ???
Yeah it's x=4 y= -1, you don't have to differentiate but it is a personal preference

EDIT: apart from it would be x-4 inside the brackets as there was a negative in the original equation
12. (Original post by maccaism)
so the minimum value is -1?
and would the x and y co ordinates be 3,5?
The minimum value is the y value, so the coordinates would be 4,-1
13. (Original post by maccaism)
yep i did completing the square and i got that answer too. i just forgot how to find the minimum value and the x and y co ordinates.

so the minimum value is -1?
and would the x and y co ordinates be 3,5?

edit: (x-4)* so x= 4, the answer is (4,-1) , not ( -4,-1)
14. (Original post by ihatehannah)
x= -4 , y=-1 , you mean.
No the original equation is x^2 -8x +15 therefore...
dy/dx = 2x -8
2x-8=0
2x=8
x=4
15. (Original post by Nirvana27)
No the original equation is x^2 -8x +15 therefore...
dy/dx = 2x -8
2x-8=0
2x=8
x=4
oh my bad, thought it was x^2 + 8x + 15, yes x= 4.
16. (Original post by ihatehannah)
the bracket ( x+4) , for it to equal to zero, x must be -4, and we know y= -1, therefore the answer is ( -4, -1)
x^2 -8x +15
(x-4)^2 -16 +15
(x-4)^2 -1
Doing it the other way
17. (Original post by Nirvana27)
x^2 -8x +15
(x-4)^2 -16 +15
(x-4)^2 -1
Doing it the other way
So the minimum value is 4,-1
x=4 and y= -1
18. (Original post by maccaism)
So the minimum value is 4,-1
x=4 and y= -1
Yeah, the minimum value is always the y value so -1, but the min coordinates are (4,-1).
19. (Original post by Nirvana27)
Yeah, the minimum value is always the y value so -1, but the min coordinates are (4,-1).
Thank you very much!
20. (Original post by maccaism)
how do i find the minimum/maximum point of this equation:

x2-8x+15

the 2 being the squared
Firstly, you will need to find gradient of the equation by calculating the derivative.
(X^2)-8X+15.......
dy/dx = 2x-8
To calculate the minima or maxima, use the calculated gradient.
set the equation equal to zero..
2X-8 = 0.
Making X as the subject, 2X-8=0
2X=8
X=8/2
X=4
as the gradient changes from -ve to +ve, for example the point reached is 4 and is positive, so you will have a minimum at x= 4.
Using the X value, substituting X=4 into y= (x^2)-8X+15
you will get y= -1.
therefore the minimum point is (4, -1) .
sorry I have plotted the graph wrong, but I hope it will give you an Idea.
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