Differentiation provides a way to get an approximation for gradient at a given point.
Consider the curve
y= x^2 +6x + 8
Differentiating to check gradient at points is:
dy/dx = 2x + 6
This is the gradient machine, can now find gradient for any point
What is the gradient at a turning point?
It is 0. Explain why?
At what point is this curve turning
when gradient = 0
2x + 6 = 0
so x = -3
curve turns when x = -3
What is the coordinate for this turning point?
when x = -3, what is y
substitute x =-3 into the curve y= x^2 +6x + 8
Show that the co-ordinate is (-3, 1)
Now take the same curve y= x^2 +6x + 8
and complete the square into the form a(x +b)^2 + c
show that the turning point is -3 and that the curve is a minimum with a value of 1.
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y=x^2+6x+8
ReplyDeletedy/dx=2x+6
d2y/dx2=2
it is not 0.
it is 2 which is positive and a min point.
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ReplyDeleteFor a function: y = f(x), a stationary point is a point on the function graph where the gradient of the function is zero...... to determine if the turning point / gradient is zero we differentiate....but for this solution dy/ dx is not equal to zero...
ReplyDeleteRemember that the turning point is where dy/dx, or the gradient, is equal to zero.
ReplyDeleteturning point: dy/dx = 0
So if we have dy/dx = 2x + 6, the turning point is where that is equal to zero, right?
Thus! dy/dx = 2x + 6 = 0
So, we can then say that the turning point is:
0 = 2x + 6
-6 = 2x
2x = -6
x = -3
So, where x = -3, the turning point is 0.
As Miss said, then we just substitute x = -3 into the original equation to find y.
y = x^2 +6x + 8
y = -3^2 + [6 (-3)] + 8
y = 9 - 18 + 8
y = 1
This comment has been removed by the author.
ReplyDeleteax² + bx = -c
ReplyDeleteWe then take 1/2b, and square it, and add it to both sides:
ax² + bx + (1/2b)² = -c + (1/2b)²
Then put the left side in square form
a(x + 1/2b)² = -c + (1/2b)²
Can you continue from there to put it in the form a(x+b)² + c, using the values given?
the turning point is = o because anytime it turns it goes to a horizontal line....and horizontal lines has a gradient of 0....becuze there is no change in y
ReplyDeleteThis comment has been removed by the author.
ReplyDeletefrom "At what point is this curve turning
ReplyDeletewhen gradient = 0
2x + 6 = 0
so x = -3
curve turns when x = -3
What is the coordinate for this turning point?"
then y=x^2 + 6x
then y=(-3)^2 + 6(-3)+8
so y = 9 + (-18) +
therefore y = -9+ 8
there for the coordinate is : (-3,-1) and not (-3,1)
at stationery pts is zero because at the max or min where the curve turns is parallel to the x axis and parrallel lines have same gradient and because the x axis is straight and at runs at a angle of 180 degrees and does not change it would be zero therfore making stat. pts equal to zero.
ReplyDeletethe gradient at the turning point is not zero it is 2. because when the first differential is it brings it to the from of:
ReplyDeletey= mx+c
where m being the gradient.
when x= (-3)
ReplyDeletey= (-3)^2+6(-3)+8
= 9-18+8
= -1
so the turning pts. are (-3,-1).
the value for y is not 1 it is -1.
ReplyDeletewhen x= -3
y= (-3)^2+6(-3)+8
= 9-18+8
= -1
x^2+6x+8
ReplyDelete(x+3)^2
so the turning pt. is -3.
(x+3)^2
=x^2+6x+6
so you need 2 more to make 8
therefore (x+3)^2+2
when x= -3
y= (-3)^2+6(3)+8
= 9-18+8
= -1
so therefore the curve is not a mim., it is a max.because it has a value of -1. and a negative value gives a max pt.
the coordinates are (0,-3)
ReplyDeletethe gradient = 0, use differentiation to solve....since the whole equation will be equal to 0.dy/dx=0
Gradient at a turning point is zero because.....well....ammmmm....lets use an example, say you have a curve increasing or going up, it then 'stops' and decreases or goes down. The point at which it stops is on a horizontal line or flat plane or parallel to the x axis, however you want to put it. Further on, a horizontal line has no elevation in other words no gradient.. hence the turning point of a curve is zero.
ReplyDeleteThe curve will turn at the point (-3,0)
ReplyDeleteX --> -3 because when gradient = 0
2x + 6 = 0
so x = -3
y --> 0 because there is no elevation at this point and no elevation means no y value.
considering the curve
ReplyDeletey=x^2+6x+8
dy/dx=2x+6
d2y/dx2=2
it is 2 which means that it is positive and has a min point.
Remember that the turning point is where dy/dx, or the gradient, is equal to zero.
ReplyDeleteturning point: dy/dx = 0
So if we have dy/dx = 2x + 6, the turning point is where that is equal to zero, right?
Thus! dy/dx = 2x + 6 = 0
So, we can then say that the turning point is:
0 = 2x + 6
-6 = 2x
2x = -6
x = -3
So, where x = -3, the turning point is 0.
As Miss said, then we just substitute x = -3 into the original equation to find y.
y = x^2 +6x + 8
y = -3^2 + [6 (-3)] + 8
y = 9 - 18 + 8
y = 1
The gradient at a turning point is zero because it is a stationary point on the graph. It is the point at which the graph changes direction, whether positive or negative.
ReplyDeletey = x^2 + 6x + 8
ReplyDelete= (x^2 + 6x) + 8
= (x + 3)^2 + 8
[ (x+3)^2 = x^2 + 6x + 6]
requires 2 more to make 8
= (x + 3)^2 + 2
finding turning point:
(x + 3) ^2 = 0
x = -3
sub x = -3 into equation:
y = x^2 + 6x + 8
y = (-3)^2 + 6(-3)+ 8
y = 9 - 18 + 8
y = 9 - 10
y = -1
The turning point is (-3,-1)
being a negative value, the turning point is a maximum.
y=x^2+6x+8
ReplyDeletedy/dx=2x+6
d2y/dx2=2
is not = 0.
it is 2 which is positive and a min point.
The turning point is the point at which the graph changes direction (usually in opposite direction). to change from positive to negative the graph must 'pass through 0'. the point at which the graph 'passes through 0' or has a gradient of zero is the turning point
ReplyDeletey= x^2 +6x + 8
ReplyDeletex= -3
y= (-3)^2 +6(-3) + 8
point at which graph turns is (-3, -1)
Gradient at a turning point is zero because it is a stationary point on the graph..
ReplyDeletehorizontal lines have no gradient. therefore dy/dx = 0
ReplyDeletein curves,gradient of a tangent at a point gives us the gradient of the curve at that point.at turning points, we say that the gradient is = 0 because at that point, a plateu of sorts is reached, i.e., at that point, the tangent to the curve is horizontal and thus has no gradient.
differenciate
ReplyDeletex^5/5 + x^4/4 + x^3/3 + x^2/2
Differenciation question
ReplyDeleteIf W=3r^3 - 2r^2 - r -3
find dW/dr
If s= 3/4-2r find ds/dr and the values of s when ds/dr= 1/6
ReplyDeletethe gradient of a straight line is constant.It is equal to the ratio y-step/x-step between any two points of the line
ReplyDeleteOn a curve the gradient is changing from one point to another.We define the gradient at any point on a curve therefore to be the gradient of the tangent to the curve at that point
ReplyDeleteif y= 16x+1/x^2, find dy/dx from first principles
ReplyDeletefind the value of x where the gradient is 0
find d^2y/dx^2
ReplyDeletea. y= 2x^3-3x^2+1
b. y= (4x-1)^3
now find (dy/dx)^2
y=x^2+6x+8
ReplyDeletedy/dx=2x+6
d2y/dx2=2
is not = 0.
it is 2 which is positive and a min point.
y = x^2 + 6x + 8
ReplyDelete= (x^2 + 6x) + 8
= (x + 3)^2 + 8
[ (x+3)^2 = x^2 + 6x + 6]
requires 2 more to make 8
= (x + 3)^2 + 2
finding turning point:
(x + 3) ^2 = 0
x = -3
sub x = -3 into equation:
y = x^2 + 6x + 8
y = (-3)^2 + 6(-3)+ 8
y = 9 - 18 + 8
y = 9 - 10
y = -1
The turning point is (-3,-1)
being a negative value, the turning point is a maximum.
y=x^2+6x+8
ReplyDeletedy/dx=2x+6
d2y/dx2=2
is not = 0.
it is 2 which is positive and a min point.
x^2+6x+8
ReplyDelete(x+3)^2
so the turning pt. is -3.
(x+3)^2
=x^2+6x+6
so you need 2 more to make 8
therefore (x+3)^2+2
when x= -3
y= (-3)^2+6(3)+8
= 9-18+8
= -1
so therefore the curve is not a mim., it is a max.because it has a value of -1. and a negative value gives a max pt.
at the turning point, the gradient is 0 because it is neither increasing nor decreasing.
ReplyDeleteit is momentarily constant as it turns.
The turning point is the point at which the graph changes direction (usually in opposite direction). to change from positive to negative the graph must 'pass through 0'. the point at which the graph 'passes through 0' or has a gradient of zero is the turning point
ReplyDeletefor the given equation of the curve,
ReplyDeleteuse the completing the square, method,
extract the turning point.
substitute back into the eqn to find y co-ordinate
and plot the curve, and determine if it is maximum or minimum.