This is a basic line thus, any parallel line also has the same equation, wince it passes through the origin it must beOriginally Posted by dadon
Sorry for so many questions. Im just stuck on these and they are meant to be simple!
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1) Find the equation of the line through the origin which is parallel to the line 4x - 27 - 5 = 0
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2) The equation of a curve is y = (x-3)(x+4). Find the gradient of the curve:
a) at the point the curve crosses the y-axis
b) at each of the points where the curve crosses the x-axis
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3) Find the coordinates of the point(s) on the curve y=3x^3 - x + 8 at which the gradient is
a) 8
b) 0
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4) The equation of a curve s=4t^2 +5t. Find the gradient of the normal at each of the points where the curve crosses the t-axis.
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5) If the gradient of a curve is y=2x^2 -3x -2. Find
a) the gradient at the point where x=0
b) the co-ordinates of the point where y=0
c) the gradient at each of the points found in b)
You probably have a typo here, given the relative location of the y and 7Originally Posted by dadon
keys I will guess this should have been:
" Find the equation of the line through the origin which is parallel to the line 4x - 2y - 5 = 0"
This may be rewritten:
,
and hence has slope , and so the equation through the origin
parallel to this also has slope and so has equation:
.
RonL
Expanding the terms on the RHS and then differentiation gives:Originally Posted by dadon
,
so:
a)The curve crosses the y axis when , hence the gradient
of the curve is obtained by plugging this value of into the expression
for the derivative and so the required gradient is
b)The curve crosses the x axis at and .
Plugging these into the expression for the derivative give gradients of and respectively.
RonL
The derivative of is:Originally Posted by dadon
a) Now this is when:
,
or:
or , which may be plugged into the equation for y to
give when and when .
So the required points are and .
b)Now this is when:
,
or:
or .
Plugging these values back into the equation for gives the other
coordinate, and so we find the required points are: and .
RonL
The points at which the curve crosses the t-axis correspond to the roots of:Originally Posted by dadon
which are and .
Now:
,
so the slopes at the two points are and respectively.
Now the gradient of the normal is -1 over the gradient of the tangent, and so
the gradients of the two normals are and respectively.
RonL
Sorry the question should be:This has a confusing use of y for the gradient of the curve, and assuming this is right there is sufficient information to find only the x-coords for the answer to part b), and c) is redundant as it is given as 0.
5) If the equation of a curve is y=2x^2 -3x -2. Find
a) the gradient at the point where x=0
b) the co-ordinates of the point where y=0
c) the gradient at each of the points found in b)