and comment on your observations
Find the coordinate of the turning point and determine whether it is a maximum or a minimum
and what is its maximum or minimum value.
- y = x^2 + x -6
- y = x^2 + 14x -5
- y = x^2 -23x + 18
- y = 2x^2 - 8x + 5
- y = 5x^2 -30x + 12
- y = 8x^2 -32x + 25
1) dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
it is positive so it is a min point..
2) dy/dx=2x+14
ReplyDeleted2y/dx2= 2
it is positive so it is a min point.
dy/dx=2x-23
ReplyDeleted2y/dx2=2
it is positive and therefor a min point.
4)dy/dx=2(2x)-8
ReplyDeletedy/dx=4x-8
d2y/dx2= 4
it is positive so a min point..
5)dy/dx=2(5x)-30
ReplyDeletedy/dx=10x-30
d2y/dx2= 10
it is positive and therefore a min point.
6) dy/dx=2(8x)-32
ReplyDeletedy/dx=16x-32
d2y/dx2=16
it is positive and therefore a min point.
to find if a point is a max or min you have to differenciate twice to see if it is positive or negative.if positive it is a minium point and if negative it is a max point.
ReplyDeletea turning point can be found when differentiating and a term is equal to zero.
ReplyDeleteboth completing the square and differentiation can give you values to plug in and plot a graph
ReplyDeleteDifferentiation is what you do to a curve to obtain its gradient
ReplyDeleteThis comment has been removed by the author.
ReplyDeletecompleting the square has real roots
ReplyDelete1)y=x^2 + x-6
ReplyDeleteby completing the square
x^2 + x-6=0
x^2 + x=6
x^2 + x+(1/2x)^2= 6+(1/2)^2
(x+ 1/2)^2= 6+ 1/4
x+ 1/2= +/- sqrt 6 1/4
so
x = -1/2+ sqrt 6 1/4
or
x = -1/2 - sqrt 6 1/4
to find out if a point is maximum or minimum, you have to differenciate twice to arrive at an answer.
ReplyDeleteif (+)-ive it is a minimum point
if (-)-ive it is a maximum point
1) y = x^2 + x - 6
ReplyDeletecompleting the square approach
y = x^2 + x -6
(x + 0.5)^2 - 0.25 - 6 = 0
(x + 0.5)^2 - 6.25
6.25 = x + 0.5
+/- 2.5 = x + 0.5
x = 2.5 - 0.5 = 2
OR
x = -2.5 - 0.5 = -3
2)y = x^2 + 14x - 5
ReplyDeletecompleting the square approach
y = x^2 + 14x - 5
(x + 7)^2 - 49 -5 = 0
(x + 7 )^2 - 54 = 0
54 = (x + 7)^2
+/- 7.35 = x + 7
7.35 - 7 =0.35
OR
-7.35 - 7 = -14.35
3) y = x^2 - 23x + 18
ReplyDeletecompleting the square approach
y = x^2 - 23x + 18
(x - 11.5)^2 - 132.25 + 18 = 0
(x - 11.25)^2 - 114.25 = 0
114.25 = (x - 11.25)^2
+/- 10.69 = x - 11.5
10.69 + 11.5 = 22.189
OR
-10.69 + 11.5 = 0.81
how do you know when it is max or min????
ReplyDeletefind the coordinates of the stationary point of the curve y=x^2+16/x determine the nature of the point
ReplyDeletey=x^2+16/x
y=X^2+16x^-1
dy/dx=2x-16x^-2
=2x-16/x^2=0
*byx^2 = 2X^3-16=0
2x^3=16
x^3=8
x=2
y=12
dy/dx=2x-16x^-2
=2+32^-3
=2+32/x^3
=6 at the pt (2,12) postive value min turning point
1) dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
it is + so it is a minimum point
2)y = x^2 + 14x - 5
y = x^2 + 14x - 5
(x + 7)^2 - 49 -5 = 0
(x + 7 )^2 - 54 = 0
54 = (x + 7)^2
+/- 7.35 = x + 7
y=x^2+x-6
ReplyDeletedy/dx 2x+1
from what i understand of differentiation you first take the formulay2-y1/x2-x1(the formula for finding rate of change on a straight lined graph)then apply it to a curved lined graph using the limiting processand then simplify the equation which gives you a general formula.from studying the the differences in the original and changede equation you obtain a general set of shortcut rules for differentiating terms anX^n-1
which i gues is what i did.
1). dy/dx= 2x+1
ReplyDeleted2y/dx2= 2
it has a mim.pt.
2). dy/dx= 2x+14
ReplyDeleted2y/dx2= 2
it has a mim. pt.
3). dy/dx= 2x-23
ReplyDeleted2y/dx2= 2
it has a mim. pt.
4). dy/dx= 4x-8
ReplyDeleted2y/dx2= 4
it has a mim. pt.
5). dy/dx= 10x-30
ReplyDeleted2y/dx2= 10
it has a mim. pt.
6). dy/dx= 16x-32
ReplyDeleted2y/dx2= 16
it has a mim. pt.
when finding the mim. or max. pt you have to differentiate twice. If it has a positive value it is mim. and if it has a negative value it is max.
ReplyDeletewhen the first differential is done you could find the gradient and the turning pts. also when completing the square you could find the turning points of the curve.
ReplyDelete1.x^2+ x^-6
ReplyDeletedy/dx= 2x+1
dy/dx= 2
min. pt = +2
to swag..
when it turns one must be a max. and one must be a min. (+ve or -ve)
min. -ve
max. +ve
ques 6
ReplyDeletecompleting the square
y=8x^2-32x+25
8(x^2-4x+25/8)
8{x^2-4x+(2)^2+25/8-(2)^2}
8{(x+2)^2+25/8-4}
8{(x+2)^2 -7/8}
8(x+2)^2 (-7/8 *8)
8(x+2)^2 -7
it has a min pt
ques 4
ReplyDeletecompleting the square
y= 2x^2 - 8x + 5
2(x^2 - 4x) + 5
2(x^2 -4x - 2^2) + 5 - 2(-2^2)
2(x - 2)^2 + 5 - 8
2(x - 2)^2 - 3
Turning point at (2,-3)
It has a Min pt
ques 4
ReplyDeletedifferentiation
y = 2x^2 -8x +5
dy/dx = 4x - 8
d2y/dx^2 = 4
It is positive so it is a min pt.
Differentiation is faster than completing the square but it does not provide the coordinates for the turning point it only gives the nature of the turning point.
The stationary or turning point is determined by differentiating a second time where the max(-ve) and the min(+ve) values can are found.
ReplyDeletedy/dx = 0
ReplyDeletetherefor
the gradiant = 0
dy/dx (this is called a turning point or a stationary point
ReplyDeleted^2y/dx^2 (this means differentiate a second time).
ReplyDeleteThe turning point in a graph is where the gradient is = 0.
ReplyDeletewhen u differentiate 2times, the value attained can be used to determine whether the turning point is max or min
ReplyDeleteto calculate the coordinates of this point, the value found after the second diff can be used as x
ReplyDeletewhen using this value as x, substitute it into the equation of the graph to find the value for y
ReplyDelete1. y = x^2 + x -6
ReplyDeletedy/dx = 2x + 1
d2y/dx^2 = 2
2>0 therefore point is min
using x=2, y=(2)^2 + 2 - 6
ReplyDeletey= 0
turning point occurs at (0, 2)
at this point grad = 0
2. y = x^2 + 14x -5
ReplyDeletedy/dx = 2x + 14
d2y/dx^2 = 2
2>0 therefore point is min
using x=2, y=(2)^2 + 14(2) - 5
ReplyDeletey= 27
turning point occurs at (27, 2)
at this point grad = 0
y = x^2 -23x + 18
ReplyDeletedy/dx= 2x -23
d2y/dx^2= 2
2>0, turning point is min
using x=2, y=(2)^2 - 23(2) +18
ReplyDeletey= -24
turning point occurs at (-24, 2)
at this point grad = 0
y = 2x^2 - 8x + 5
ReplyDeletedy/dx = 4x - 8
d2y/dx^2 = 4
4>0 turning point is min
using x=4, y=2(4)^2 - 8(4) +5
ReplyDeletey= 5
turning point occurs at (5, 4)
at this point grad = 0
y = 5x^2 -30x + 12
ReplyDeletedy/dx = 10x -30
d2y/dx^2 = 10
10>0, turning point is min
using x=10, y=5(10)^2 - 30(10) +12
ReplyDeletey= 212
turning point occurs at (212, 10)
at this point grad = 0
y = 8x^2 -32x + 25
ReplyDeletedy/dx = 16x - 32
d2y/dx^2 = 16
16>0, turning point is min
using x=16, y= 8(16)^2 - 32(16) +25
ReplyDeletey= 1561
turning point occurs at (1561, 16)
at this point grad = 0
the turning point of a graph is when the gradient is equal to zero
ReplyDelete1. x^2+x^-6
ReplyDeletethen dy/dx=2x+1
dy/dx=+2
therefore min point=+2
an when a graph has a turning point there must be a max. or min. value
max. is minus
min. is postive
oh an that was for swag
ReplyDelete4.dy/dx= 4x-8
ReplyDeleted2y/dx2=4
therefore it is a min point.=+2
2.dy/dx=2x+14
ReplyDeleted2y/dx2=2
therfore it is a min. point=+2
6.dy/dx=2(8x)-32
ReplyDeletedy/dx=16x-32
d2y/dx2=16
therefore is a min. point=+16
3. dy/dx=2x-23
ReplyDeleted2y/dx2=2
therefore it is a min. point=+2
4)dy/dx=2(2x)-8
ReplyDeletedy/dx=4x-8
d2y/dx2= 4
positive therefore it is a min point..
2. y = x^2 + 14x -5
ReplyDeletedy/dx = 2x + 14
d2y/dx^2 = 2
2>0 therefore point is min
5)dy/dx=2(5x)-30
ReplyDeletedy/dx=10x-30
d2y/dx2= 10
positive resulting in a min point.
3. dy/dx=2x-23
ReplyDeleted2y/dx2=2
positive therefore it is a min
1) dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
it is positive so it is a min point.
2) dy/dx=2x+14
ReplyDeleted2y/dx2= 2
it is positive, therefore is a min point.
3) y = x^2 - 23x + 18
ReplyDeleteBy completing the square.....
y = x^2 - 23x + 18
(x - 11.5)^2 - 132.25 + 18 = 0
(x - 11.25)^2 - 114.25 = 0
114.25 = (x - 11.25)^2
+/- 10.69 = x - 11.5
10.69 + 11.5 = 22.189
4)dy/dx=2(2x)-8
ReplyDeletedy/dx=4x-8
d2y/dx2= 4
it is positive and therefore has a min point.
5)dy/dx=2(5x)-30
ReplyDeletedy/dx=10x-30
d2y/dx2=10
it is positive and therefore has a min point.
In reply to swag's question........
ReplyDeleteits min when its positive and its max when its negative!!!!!!!!
1) dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
it is positive so it is a min point..
quest. 6
ReplyDeletedy/dx=2(8x)-32
dy/dx=16x-32
d2y/dx2=16
therefore is a min. point
1. y = x^2 + x - 6
ReplyDeletedy/dx= 2x + 1
d2y/dx2=2
d2y/dx2 is positive, therefore it is a minimum
2. y = x^2 + 14 x - 5
ReplyDeletedy/dx=2x+14
d2y/dx2= 2
d2y/dx2 is positive therefore it is a minimum
3. y = x^2 -23x + 18
ReplyDeletedy/dx = 2x - 23
d2y/dx2 = 2
d2y/dx2 is positive, therefore it is a minimum
when the gradient is zero that is the turning point in the graph
ReplyDeleteThe turning point is found by differentiating a second time and the max(-ve) and the min(+ve) values can are found.
ReplyDeletethe turning point of a graph is when the gradient is equal to zero
ReplyDelete1. y = x^2 + x -6
ReplyDeletedy/dx = 2x + 1
d2y/dx^2 = 2
d2y/dx^2=2 therefore point is min
3).y = x^2 -23x + 18
ReplyDeletedy/dx= 2x-23
d2y/dx2= 2
d2y/dx2>0 therefore it is a mim. pt.
2)y = x^2 + 14x -5
ReplyDeletedy/dx=2x+14
d2y/dx2= 2
d2y/dx2>0 it is a min point.
4. y = 2x^2 -8x +5
ReplyDeletedy/dx = 4x - 8
d2y/dx^2 = 4
d2y/dx2>0 it is a min point
5)y = 5x^2 -30x + 12
ReplyDeletedy/dx=10x-30
d2y/dx2= 10
d2y/dx2>0 it is a min point
6).y = 8x^2 -32x + 25
ReplyDeletedy/dx= 16x-32
d2y/dx2= 16
d2y/dx2>0 it is a min point
2)
ReplyDeletedy/dx = 2x+14
d2y/dx2 = 2
it is positive so it is a min point
6)
ReplyDeletedy/dx = 2(8x)-32
dy/dx = 16x-32
d2y/dx2=16
it is positive and therefore a min point.
dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
it is positive so it is a min point
dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
2 positive so it is a min point
4)
ReplyDeletedy/dx=2(2x)-8
dy/dx=4x-8
d2y/dx2= 4
it is positive so a min point
(1)y = x^2 + x -6
ReplyDeletedy/dx= (2*1)x^(2-1) + x^(1-1)
dy/dx= 2x + 1
sub. dy/dx=0
0 = 2x +1
2x= -1
x= -1/2
(1)sub x= -1/2 into y= x^2 + x -6
ReplyDeletey= (-1/2)^2 + (-1/2) -6
y= 1/4 -1/2 - 6
y=-1/4 - 6
y=-25/4
therefore the the coordinates of the turning point are, x= -1/2 , y= -25/4
(1)since dy/dx= 2x-1
ReplyDeletein order to determine if it is a max or a min u have to differentiate again.
d2y/dx^2= 2
since the 2 is more than o it is a minimum value
number1
ReplyDeletey = x^2 + x -6
dy/dx= 2x + 1
d2y/dx2=2
it is positive so it is a min point.
nummber3
ReplyDeletey = x^2 -23x + 18
dy/dx= 2x-23
d2y/dx2= 2
d2y/dx2>0 therefore it is a min pt.
looking at d the above ans...
ReplyDeletei can determine that the 2nd differential, can be used to determine whether the turning point is max or min.
if the value is:
ReplyDeletepositive= min pt
negative= max pt
number6
ReplyDeletey = 8x^2 -32x + 25
dy/dx=2(8x)-32
dy/dx=16x-32
d2y/dx2=16
therefore is a min.
number5
ReplyDeletey = 5x^2 -30x + 12
dy/dx=2(5x)-30
=10x-30
d^2y/dx^2=10
it is positive and therefore has a min point.
i want to correct a mistake i made when i was typing the second differential. i forgot to put the powers. and that can change its whole meaning. for numbers 1, 3, and 6
ReplyDeletey=8x^2-32x+25
ReplyDeletedy/dx= 2(8x)^2-1 - 32
dy/dx= 16x - 32
d^2y/dx^2 = 16
16 is +ve therfore a min turning pt exists
PS
ReplyDeletethat was the ans to no. 6 by the way
5)
ReplyDeletey = 5x^2 - 30x +12
dy/dx = 10x -30
d^2y/dx^2 = 10
a min turning pt exists
the equation for a curve is
ReplyDeleteax^2 + bx +c
will it be right to say that if
a is a positive no. a minimum turning point will exist.
????
4)
ReplyDeletey = 2x^2 -8x + 5
dy/dx= 4x-8
d^2y/dx^2 = 4
min tp exists
completing the square approach
ReplyDeletequestion 1
a=1, b=1, c=-6
h= b/2a
= 1/2(1)
= 1/2
k= 4ac - b^2/4a
= 4(1)(-6) - (1)^2/4(1)
= -24-1/4
= -25/4
= -6.25
a(x+h)^2 + c
(x+ 1/2)^2- 6.25
minimum value is -6.25
turning point (-1/2, -6.25)
differentiation approach
ReplyDeletedy/dx = 2x + 1
d2y/dx2 = 2
turning point is a minimum
completing the square approach
ReplyDeletequestion 2
a=1, b=14, c=-5
h= b/2a
= 14/2(1)
= 14/2
=7
k= 4ac - b^2/4a
= 4(1)(-5) - (14)^2/4(1)
= -20-196/4
= -216/4
= -54
a(x+h)^2 + c
(x+ 7)^2- 54
minimum value is -54
turning point (-7,-54)
differentiation approach
ReplyDeletedy/dx = 2x+14
d2y/dx2 = 2
turning point is a minimum
completing the square approach
ReplyDeletequestion 3
a=1, b=-23, c=18
h= b/2a
= -23/2(1)
= -11.5
k= 4ac - b^2/4a
= 4(1)(18) - (-23)^2/4(1)
= 72-529/4
=-457/4
=-114.25
a(x+h)^2 + c
(x-11.5)^2- 114.25
minimum value is -114.25
turning point (11.5, -114.25)
differentiation approach
ReplyDeletedy/dx = 2x-23
d2y/dx2 = 2
turning point is minimum
This comment has been removed by the author.
ReplyDeletedifferentiation approach
ReplyDeletedy/dx = 4x-8
d2y/dx2 = 4
turning point is a minimum
(2)y = x^2 + 14x -5
ReplyDeletedy/dx= (2*1)x^(2-1) + 14x^(1-1)
dy/dx= 2x + 14
sub dy/dx=0
2x+ 14=0
2x=-14
x= -14/2
x= -7
sub x= -7 into y= x^2 + 14x -5
ReplyDeletey= (-7)^2 + 14(-7) -5
y= 49 - 98 -5
y= -54
therefore the coordinates of the turning point are ; x= -7, y= -54
This comment has been removed by the author.
ReplyDelete(2)since dy/dx= 2x + 14
ReplyDeletein order to determine if it is a max or a min u have to differentiate again.
d2y/dx^2= 2
since the 2 is more than 0 it is a minimum value
(6)y = 8x^2 -32x + 25
ReplyDeletedy/dx= (8*2)x^(2-1) - 32x^(1-1)
dy/dx= 16x - 32
sub dy/dx=0
16x- 32=0
16x= 32
x= 32/16
x= 2
(6) sub x=2 into y = 8x^2 -32x + 25
ReplyDeletey= 8(2)^2 - 32(2) + 25
y= 32 - 64 + 25
y= 57 - 64
y= -7
therefore the coordinates of the turning point are ; x=2, y=-7
(6)since dy/dx= 16x -32
ReplyDeletein order to determine if it is a max or a min u have to differentiate again.
d2y/dx^2= 16
since the 16 is more than 0 it is a minimum value
1) dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
min point.
y = x^2 + x -6
ReplyDelete(x + 0.5)^2 - 0.25 - 6 = 0
(x + 0.5)^2 - 6.25
6.25 = x + 0.5
+/- 2.5 = x + 0.5
x = 2.5 - 0.5 = 2
2) dy/dx=2x+14
ReplyDeleted2y/dx2= 2
min point.
3)dy/dx=2x-23
ReplyDeleted2y/dx2=2
min point.
4)dy/dx=4x-8
ReplyDeleted2y/dx2= 4
min point
5)dy/dx=10x-30
ReplyDeleted2y/dx2= 10
min point.
6) dy/dx=16x-32
ReplyDeleted2y/dx2=16
min point.
This comment has been removed by the author.
ReplyDeletedifferentiation approach
ReplyDeletedy/dx = 10x-30
d2y/dx2 = 10
turning point is a minimum
completing the square approach
ReplyDeletequestion 4
a=2, b=-8, c=5
h= b/2a
= -8/2(2)
= -8/4
= -2
k= 4ac - b^2/4a
= 4(2)(5) - (-8)^2/4(2)
= -24/8
= -3
a(x+h)^2 + c
2(x-2)^2- 3
minimum value is -3
turning point (2,-3)
completing the square .
ReplyDeletequestion 5
a=5, b=-30, c=12
h= b/2a
= -30/2(5)
= -30/10
= -3
k= 4ac - b^2/4a
= 4(5)(12) - (-30)^2/4(5)
= -240-900/20
=-660/20
= -33
a(x+h)^2 + c
5(x-3)^2- 33
minimum value is -33
turning point (3,-33)
completing the square approach
ReplyDeletequestion 6
a=8, b=-32, c=25
h= b/2a
= -32/2(8)
= -32/16
= -2
k= 4ac - b^2/4a
= 4(8)(25) - (-32)^2/4(8)
= 800-1024/32
= -224/32
= -7
a(x+h)^2 + c
8(x-2)^2- 7
minimum value is -7
turning point (2,-7)
differentiation approach
ReplyDeletedy/dx = 16x-32
d2y/dx2= 16
turning point is a minimum
use differentiation and the completing the square approach
ReplyDeleteFind the coordinate of the turning point and determine whether it is a maximum or a minimum
and what is its maximum or minimum value.
y=x^2 + 6x + 8
using both methods work
ReplyDeletey=4x^2 - 22x + 24
using both methods work
ReplyDeletey=3x^2 - 10x + 7
using both methods work
ReplyDeletey=6x^2 - 7x -20
using both methods work
ReplyDeletey=2x^2 + 11x + 12
1)
ReplyDeletedy/dx= 2x + 1
d2y/dx2=2
it is positive so it is a min point..
2)
ReplyDeletedy/dx=2x+14
d2y/dx2= 2
it is positive so it is a min point.
3.
ReplyDeletedy/dx=2x-23
d2y/dx2=2
it is positive and therefor a min point
4)
ReplyDeletedy/dx=2(2x)-8
dy/dx=4x-8
d2y/dx2= 4
it is positive so a min point.
5)
ReplyDeletedy/dx=2(5x)-30
dy/dx=10x-30
d2y/dx2= 10
it is positive and therefore a min point.
6)
ReplyDeletedy/dx=2(8x)-32
dy/dx=16x-32
d2y/dx2=16
it is positive and therefore a min point.
1) dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
it is positive so it is a min point..
2) dy/dx=2x+14
ReplyDeleted2y/dx2= 2
it is positive so it is a min point.
dy/dx=2x-23
ReplyDeleted2y/dx2=2
it is positive and therefor a min point.
4)dy/dx=2(2x)-8
ReplyDeletedy/dx=4x-8
d2y/dx2= 4
it is positive so a min point
5)dy/dx=2(5x)-30
ReplyDeletedy/dx=10x-30
d2y/dx2= 10
it is positive and therefore a min point.
6) dy/dx=2(8x)-32
ReplyDeletedy/dx=16x-32
d2y/dx2=16
it is positive and therefore a min point.
to find if a point is a max or min you have to differenciate twice to see if it is positive or negative.if positive it is a minium point and if negative it is a max point.
ReplyDelete1. y = x^2 + x -6
ReplyDeletedy/dx = 2x + 1
d2y/dx2 = 2
since the value of the second derivative is positive, it is a minimum point
2. y = x^2 + 14x -5
ReplyDeletedy/dx = 2x + 14
d2y/dx2 = 2
since the value of the second derivative is positive, it is a minimum point
3. y = x^2 -23x + 18
ReplyDeletedy/dx = 2x - 23
d2y/dx2 = 2
since the value of the second derivative is positive, it is a minimum point
4. y = 2x^2 - 8x + 5
ReplyDeletedy/dx = 4x - 8
d2y/dx2 = 4
since the value of the second derivative is positive, it is a minimum point
5. y = 5x^2 -30x + 12
ReplyDeletedy/dx = 10x - 30
d2y/dx2 = 10
since the value of the second derivative is positive, it is a minimum point
6. y = 8x^2 -32x + 25
ReplyDeletedy/dx = 16x - 32
d2y/dx2 = 16
since the value of the second derivative is positive, it is a minimum point
If the value of the second derivative was negative, then it would be a maximum point
ReplyDelete1) dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
it is positive so it is a min point..
2) dy/dx=2x+14
ReplyDeleted2y/dx2= 2
it is positive so it is a min point.
3) dy/dx=2x-23
ReplyDeleted2y/dx2=2
it is positive and therefor a min point.
4)dy/dx=2(2x)-8
ReplyDeletedy/dx=4x-8
d2y/dx2= 4
it is positive so a min point..
5)dy/dx=2(5x)-30
ReplyDeletedy/dx=10x-30
d2y/dx2= 10
it is positive and therefore a min point.
6) dy/dx=2(8x)-32
ReplyDeletedy/dx=16x-32
d2y/dx2=16
it is positive and therefore a min point.
1.dy/dx= 2x + 1
ReplyDeleted2y/dx2=2
it is positive so it is a min point..
4.dy/dx=2(2x)-8
ReplyDeletedy/dx=4x-8
d2y/dx2= 4
it is positive so a min point.
6.dy/dx=2(8x)-32
ReplyDeletedy/dx=16x-32
d2y/dx2=16
it is positive and therefore a min point
y = x^2 + x -6
ReplyDeletedy/dx= 2x + 1
d2y/dx2=2
since it is positive, the graph has a minimum point.
y = x^2 + 14x -5
ReplyDeletedy/dx=2x+14
d2y/dx2= 2
since it is positive, the graph has a minimum point.
y = x^2 -23x + 18
ReplyDeletedy/dx=2x-23
d2y/dx2=2
since it is positive, the graph has a minimum point.
2x^2 - 8x + 5
ReplyDeletedy/dx=2(2x)-8
dy/dx=4x-8
d2y/dx2= 4
this is also positive so therefore the graph has a minimum point.
y = 5x^2 -30x + 12
ReplyDeletedy/dx=2(5x)-30
dy/dx=10x-30
d2y/dx2= 10
this is positive so therefore the graph has a minimum point.
y = 8x^2 -32x + 25
ReplyDeletedy/dx=2(8x)-32
dy/dx=16x-32
d2y/dx2=16
this is positive so therefore the graph has a minimum point.
hmmm completing the square i need to go get sum help fast because i am a bit confused with this topic
ReplyDeleteThis comment has been removed by the author.
ReplyDeletein the graph you determine the turning point when the gradient is equal to zero(0)
ReplyDeletenumber (1)
ReplyDeletey = x^2 + x -6
dy/dx = 2x + 1
d2y/dx^2 = 2
d2y/dx^2=2 therefore point is min
number (2)
ReplyDeletey = x^2 + 14x -5
dy/dx=2x+14
d2y/dx2= 2
d2y/dx2>0 it is a min point.
number (3)
ReplyDeletey = x^2 -23x + 18
dy/dx= 2x-23
d2y/dx2= 2
d2y/dx2>0 therefore it is a min. pt.
number (4)
ReplyDeletedy/dx=2(2x)-8
dy/dx=4x-8
d2y/dx2= 4
it is positive so a min point..
number (5)
ReplyDeletedy/dx=2(5x)-30
dy/dx=10x-30
d2y/dx2= 10
it is positive and therefore a min point.
number (6)
ReplyDeletedy/dx=2(8x)-32
dy/dx=16x-32
d2y/dx2=16
it is positive and therefore a min point
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteQuestion 1. SOLVED BY DIFFERECIATION
ReplyDeletey = x^2 + x -6
dy/dx= 2x+ 1
Turning point occurs at dy/dx=0
2x+1 = 0
x= -1/2
sub x= -1/2 into y = x^2 + x -6
y= (-1/2)^2+ (-1/2)-6
y= 1/4 -1/2- 6
y= -25/4
Therefore Turning point is (-1/2,-25/4)
The curve is a minimum because coefficient of x^2 is positive and the minimum value of the curve is -25/4
Question 1. SOLVED BY COMPLETEING THE SQUARE
ReplyDeletex^2 + x -6
(x+1/2)^2-6 1/4
(x+1/2)^2 – 25/4
Minimum value is -25/4 when
(x+1/2)=0
x= -1/2
Curve is a minimum, and the minimum value occurs at -25/4
Turning point is (-1/2,-25/4)
Question 2. SOLVED BY DIFFERECIATION
ReplyDeletey = x^2 + 14x -5
dy/dx= 2x+ 14
Turning point occurs at dy/dx=0
2x+14 = 0
x= -14/2
x=-7
sub x=-7 into y = x^2 + 14x -5
y= (-7)^2+ 14(-7)-5
y= 49 -98- 5
y= -54
Therefore Turning point is (-7,-54)
The curve is a minimum because coefficient of x^2 is positive and the minimum value of the curve is -54
Question 2. SOLVED BY COMPLETEING THE SQUARE
ReplyDeletex^2 + 14x -5
(x+7)^2-54
Minimum value is -54 when
(x+7)=0
x= -7
Curve is a minimum, and the minimum value occurs at -54
Turning point is (-7,-54)
Question 3. SOLVED BY DIFFERECIATION
ReplyDeletey = x^2 -23x +18
dy/dx= 2x-23
Turning point occurs at dy/dx=0
2x-23= 0
x= 23/2
sub x= 23/2 into y = x^2 -23x +18
y= (23/2)^2-23 (23/2) +18
y= 529/4 -529/2 +18
y= -457/4
Therefore Turning point is (23/2,-457/4)
The curve is a minimum because coefficient of x^2 is positive and the minimum value of the curve is -457/4
Question 3. SOLVED BY COMPLETEING THE SQUARE
ReplyDeletex^2 -23x +18
(x-23/2)^2- 114 1/4
(x-23/2)^2- 457/4
Minimum value is -457/4 when
(x-23/2)=0
x= 23/2
Curve is a minimum, and the minimum value occurs at -457/4
Turning point is (23/2,-457/4)
Question 4. SOLVED BY DIFFERECIATION
ReplyDeletey = 2x^2 -8x +5
dy/dx= 4x-8
Turning point occurs at dy/dx=0
4x-8= 0
x= 8/4
x=2
sub x=2 into y = 2x^2 -8x +5
y= 2(2)^2 -8(2) +5
y= 8-16+5
y= -3
Therefore Turning point is (2,-3)
The curve is a minimum because coefficient of x^2 is positive and the minimum value of the curve is -3
Question 4. SOLVED BY COMPLETEING THE SQUARE
ReplyDelete2x^2 -8x +5
2[x^2 -4x +5/2]
2(x-2) ^2 –(3/2x2/1)
2(x-2) ^2 –3
Minimum value is -3 when
(x-2)=0
x= 2
Curve is a minimum, and the minimum value occurs at -3
Turning point is (2,-3)
Question 5. SOLVED BY DIFFERECIATION
ReplyDeletey = 5x^2 -30x +12
dy/dx= 10x-30
Turning point occurs at dy/dx=0
10x-30= 0
x= 30/10
x=3
sub x=3 into y = 5x^2 -30x +12
y= 5(3)^2 -30(3) +12
y=45-90+12
y= -33
Therefore Turning point is (3,-33)
The curve is a minimum because coefficient of x^2 is positive and the minimum value of the curve is -33
Question 5. SOLVED BY COMPLETEING THE SQUARE
ReplyDelete5x^2 -30x +12
5[x^2 -6x +12/5]
5(x-3) ^2 –(33/5x5/1)
5(x-3) ^2 –33
Minimum value is -33 when
(x-3)=0
x= 3
Curve is a minimum, and the minimum value occurs at -33
Turning point is (3,-33)
Question 6. SOLVED BY DIFFERECIATION
ReplyDeletey = 8x^2 -32x +25
dy/dx= 16x-32
Turning point occurs at dy/dx=0
16x-32= 0
x= 32/16
x=2
sub x=2 into y = 8x^2 -32x +25
y= 8(2)^2 -32(2) +25
y= 32 – 64+25
y= -7
Therefore Turning point is (2,-7)
The curve is a minimum because coefficient of x^2 is positive and the minimum value of the curve is -7
Question 6. SOLVED BY COMPLETEING THE SQUARE
ReplyDelete8x^2 -32x +25
8[x^2 -4x +25/8]
8(x-2) ^2 –(7/8 x 8/1)
8(x-2) ^2 –7
Minimum value is -7 when
(x-2)=0
x= 2
Curve is a minimum, and the minimum value occurs at -7
Turning point is (2,-7)
for ques 4
ReplyDeletey = 2x^2 -8x +5
dy/dx = 4x - 8
d2y/dx^2 = 4
d2y/dx2>0 it is a min point
for ques 5.
ReplyDeletedy/dx=2(5x)-30
dy/dx=10x-30
d2y/dx2= 10
it is +ve and therefore a min point.
for ques 2
ReplyDeletey = x^2 + 14x -5
dy/dx=2x+14
d2y/dx2= 2
d2y/dx2>0 it is a min point.
no1 diffential
ReplyDeletey = x^2 + x -6
dy/dx = 2x +1
d2y/dx2= 2
+ve therefore min pt
no1 completing square
ReplyDeletey = x^2 + x -6
(x^2 + [(1/2)(1)]^2) -4 - 1/4
(x + 1/2)^2 - 4 1/4