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Re: can anyone help with these questions
« Reply #30 on: January 16, 2014, 12:03:54 am »
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for 3f i got the answer as y=-3/4x+55/18 but in the answers it says the answer is y=-3/4x-9/2 is the answer wrong or am i wrong ?

The latter. I'll show you the solution:

"Find the equation of the line that has gradient -3/4, passing through the intersection of the lines with equations x + 4y = −14 and −5x + 2y = 4."

y = mx + c
Since the gradient is -3/4, then m = -3/4
y = (-3/4)x + c

Now in order to work out c, you need to substitute the point (x, y) in. They have told you that the point is where the equations x + 4y = -14 and -5x + 2y = 4 intersect, so use this information to work out the point:

x + 4y = -14 ---> Equation 1
-5x + 2y = 4 ---> Equation 2

Solve the simultaneous equations:

From equation 1:
x = -14 - 4y
x = -(4y + 14)
Substitute into equation 2:
-5(-(4y + 14)) + 2y = 4
5(4y + 14) + 2y = 4
20y + 70 + 2y = 4
22y = -66
y = -3
Substitute into equation 1:
x = -14 - 4(-3)
x = -14 + 12
x = -2

So the point (x, y) = (-2, -3). Substitute this into your original equation:

y = (-3/4)x + c
If x = -2 and y = -3
-3 = (-3/4)(-2) + c
-3 = 3/2 + c
c = -3 - 3/2
c = -9/2

So the equation is:
y = (-3/4)x - 9/2
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knightrider

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Re: can anyone help with these questions
« Reply #31 on: January 16, 2014, 04:39:50 pm »
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thx for help

how would you do these questions i cant be able to figure them out

1)A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.
2)The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4,8) also lie on this line?

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Re: can anyone help with these questions
« Reply #32 on: January 16, 2014, 04:46:24 pm »
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thx for help

how would you do these questions i cant be able to figure them out

1)A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.
2)The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4,8) also lie on this line?

for question 1, you know two points for sure: (-8,-5), (4,-3). because of this you can actually find the equation of the line.
After finding the equation of the line you can sub in the point (a,12) to find the value of a. Do you get this?

and it's a similar process of 2). Find the equation of the line using the first two point. then you have the point (4,8). to test if this point lies on the line, sub it in to your new found equation. if you sub in x=4 and get y=8, and sub y=8 and get x=4. Then the point lies on the line.

Class of 2014.

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Re: can anyone help with these questions
« Reply #33 on: January 16, 2014, 04:54:10 pm »
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Can someone help me with this question? Thanks a lot

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Re: can anyone help with these questions
« Reply #34 on: January 16, 2014, 05:07:29 pm »
+1
thx for help

how would you do these questions i cant be able to figure them out

1)A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.
An altnernative approach to that of Nato would be that the gradient between each of the points must be the same i.e.



Knowing this we can equate 2 of these (if 2 are the same, then all 3 will be) to find the value of








2)The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4,8) also lie on this line?

The same again here, check whether 2 of the following hold:









Therefore it does lie on the line.

*Note: You could also just do this by inspection. The points (2,7),(4,8),(6,9) obviously follow a linear trend.

Edit:

Can someone help me with this question? Thanks a lot

You have three points on the parabola.

1. The left most point: (0,75)
2. The local minimum: (90,30)
3. The right most point (following symmetry): (180,75)

You can put these into the equation and solve for .

Alternatively you could have it in the form and sub in one of the end points.
« Last Edit: January 16, 2014, 05:15:38 pm by Phy124 »
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Re: can anyone help with these questions
« Reply #35 on: January 16, 2014, 05:19:18 pm »
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Can someone help me with this question? Thanks a lot

Hey ill do my best, but I'm a bit stuck as well and maybe if someone could tell me if I'm one the right track?

Anyway if we want to describe this graph, we can put it into turning point form a(x-b)^2+c
so the minimum is 30 as it states hence c=30, now we need to find b and the dilation a
it says that assume the parabola is the suspension cable but we can see the parabola does not touch not cross through the x intercepts meaning the discriminant is 1>discriminant so therefore it has no solutions so it has no x intercepts.
So now we know we have no x intercepts out turning point is (0,30) am i on the right track here?

Ok so now we have the turning pint, a and c we need to find A the dilation and this is where I'm stuck.

Tried my best.
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Re: can anyone help with these questions
« Reply #36 on: January 16, 2014, 05:38:32 pm »
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thx for help

how would you do these questions i cant be able to figure them out

1)A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.
2)The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4,8) also lie on this line?

1.)  Those coordinates are along 1 line, therefore have the same gradient.  So find the gradient of (-8,-5) (4,-3)=1/6
so gradient is 1/6.  Now get coordinates (-8,-5) and (a,12) now find the gradient of this and it has to =1/6,
so 12+5=17, so 17 divided by something needs to give us 1/6, so we find that 17 times 6=102.
now 17/102=1/6 so it works!  Not we have to find a so some thing a+8=102, so a=94, because 94+8=102
a=94

2.)  yes they do because (2,7) and (6,9) =1/2 so their gradient is 1/2 and if you find the gradient of (4,8) with (2,7) you'll find that it =1/2.
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Re: can anyone help with these questions
« Reply #37 on: January 16, 2014, 05:39:32 pm »
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Hey ill do my best, but I'm a bit stuck as well and maybe if someone could tell me if I'm one the right track?

Anyway if we want to describe this graph, we can put it into turning point form a(x-b)^2+c
so the minimum is 30 as it states hence c=30, now we need to find b and the dilation a
it says that assume the parabola is the suspension cable but we can see the parabola does not touch not cross through the x intercepts meaning the discriminant is 1>discriminant so therefore it has no solutions so it has no x intercepts.
So now we know we have no x intercepts out turning point is (0,30) am i on the right track here?

Ok so now we have the turning pint, a and c we need to find A the dilation and this is where I'm stuck.

Tried my best.

Why is C=30?
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Re: can anyone help with these questions
« Reply #38 on: January 16, 2014, 05:49:17 pm »
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Why is C=30?

Maximum and minimum is given by c in the form of a turning pint form hence c=30.
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Re: can anyone help with these questions
« Reply #39 on: January 16, 2014, 05:53:48 pm »
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So now we know we have no x intercepts out turning point is (0,30) am i on the right track here?


The numbers are a little awkward I think and I've kind of forgotten turning point form but I think you went wrong about half way.
You have three known points from the information given that lie on the graph:
(0,75) - the extreme left of the parabola
(90,30) - the turning point... the 90 comes from half way between the 180m distance provided, the 30 is stated in the question.
(180,75) - the extreme right of the parabola

turning point form:
y = a(x-h)^2 + c

so we can sub in our turning point:
y = a(x - 90)^2 + 30

then we can sub in one of our other points to find 'a', lets use (0,75)
75 = a(0 - 90)^2 + 30
expand
75 = 8100a + 30
subtract 30 from both sides
45 = 8100a
divide both sides by 8100
a = 45/8100

so your final answer in turning point form is

y = 45/8100 (x - 90)^2 + 30
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Re: can anyone help with these questions
« Reply #40 on: January 16, 2014, 05:55:38 pm »
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An altnernative approach to that of Nato would be that the gradient between each of the points must be the same i.e.



Knowing this we can equate 2 of these (if 2 are the same, then all 3 will be) to find the value of








The same again here, check whether 2 of the following hold:









Therefore it does lie on the line.

*Note: You could also just do this by inspection. The points (2,7),(4,8),(6,9) obviously follow a linear trend.

Edit:
You have three points on the parabola.

1. The left most point: (0,75)
2. The local minimum: (90,30)
3. The right most point (following symmetry): (180,75)

You can put these into the equation and solve for .

Alternatively you could have it in the form and sub in one of the end points.

Thought i was on the right track… Then saw this.
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Re: can anyone help with these questions
« Reply #41 on: January 16, 2014, 05:58:32 pm »
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The numbers are a little awkward I think and I've kind of forgotten turning point form but I think you went wrong about half way.
You have three known points from the information given that lie on the graph:
(0,75) - the extreme left of the parabola
(90,30) - the turning point... the 90 comes from half way between the 180m distance provided, the 30 is stated in the question.
(180,75) - the extreme right of the parabola

turning point form:
y = a(x-h)^2 + c

so we can sub in our turning point:
y = a(x - 90)^2 + 30

then we can sub in one of our other points to find 'a', lets use (0,75)
75 = a(0 - 90)^2 + 30
expand
75 = 8100a + 30
subtract 30 from both sides
45 = 8100a
divide both sides by 8100
a = 45/8100

so your final answer in turning point form is

y = 45/8100 (x - 90)^2 + 30

Didn't see that!  Yep turning point of x is half way of the x intercepts… Should have saw that.  But the parabola doesn't touch the x intercepts so i thought it had no x intercepts.
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hobbitle

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Re: can anyone help with these questions
« Reply #42 on: January 16, 2014, 05:59:31 pm »
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Didn't see that!  Yep turning point of x is half way of the x intercepts… Should have saw that.  But the parabola doesn't touch the x intercepts so i thought it had no x intercepts.

You're right, it doesn't have any x intercepts.
That doesn't mean that the x-coordinate of the turning point is zero, though.  They are unrelated things.
The extremities of the parabola along the x axis (0 --> 180) aren't intercepts.  They are just points along the x axis.  They aren't 'cutting' the x axis.
« Last Edit: January 16, 2014, 06:01:38 pm by hobbitle »
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Re: can anyone help with these questions
« Reply #43 on: January 16, 2014, 06:01:23 pm »
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You're right, it doesn't have any x intercepts.
That doesn't mean that the x-coordinate of the turning point is zero, though.  They are unrelated things.

but to find the x coordinate of the turning point don't you add the two x intercepts together, if there isn't any 0+0/2=0 so isn't it still 0? 
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Re: can anyone help with these questions
« Reply #44 on: January 16, 2014, 06:03:24 pm »
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but to find the x coordinate of the turning point don't you add the two x intercepts together, if there isn't any 0+0/2=0 so isn't it still 0? 
If there aren't any intercepts, it doesn't naturally follow that the value is 0. The value being zero means the intercept is at 0. So the x int is undefined as there is none
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