1. ## Equation of line & Gradient qus.

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)

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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This is a basic line $\displaystyle x=k$ thus, any parallel line also has the same equation, wince it passes through the origin it must be $\displaystyle x=0$

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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Thus, we have,
$\displaystyle y=(x-3)(x+4)=x^2+x-12$
If the gradient means derivative then,
$\displaystyle y'=2x+1$

It passes the y-axis when $\displaystyle x=0$ thus
$\displaystyle y=0^2+0-12=-12$

It passes the x-axis when $\displaystyle y=0$ thus,
$\displaystyle 0=x^2+x-12=(x-3)(x+4)$
Thus,
$\displaystyle x=\left\{ \begin{array}{c}-4\\3\end{array}\right$

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
If,
$\displaystyle y=3x^3-x+8$
Then,
$\displaystyle y'=9x^2-1$
Yuu need it to be zero thus,
$\displaystyle 0=9x^2-1$
Thus,
$\displaystyle 9x^2=1$
Thus,
$\displaystyle x^2=1/3$
Thus,
$\displaystyle x=\pm \sqrt{1/3}=\pm \frac{\sqrt{3}}{3}$

1) Find the equation of the line through the origin which is parallel to the line 4x - 27 - 5 = 0
You probably have a typo here, given the relative location of the y and 7
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:

$\displaystyle y=2x-5/2$,

and hence has slope $\displaystyle 2$, and so the equation through the origin
parallel to this also has slope $\displaystyle 2$ and so has equation:

$\displaystyle y=2x$.

RonL

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
Expanding the terms on the RHS and then differentiation gives:

$\displaystyle y=x^2+x-12$,

so:

$\displaystyle \frac{dy}{dx}=2x+1$

a)The curve crosses the y axis when $\displaystyle x=0$, hence the gradient
of the curve is obtained by plugging this value of $\displaystyle x$ into the expression
for the derivative and so the required gradient is $\displaystyle 1$

b)The curve crosses the x axis at $\displaystyle x=3$ and $\displaystyle x=-4$.
Plugging these into the expression for the derivative give gradients of $\displaystyle 7$ and $\displaystyle -7$ respectively.

RonL

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
The derivative of $\displaystyle y=3x^3-x+8$ is:

$\displaystyle \frac{dy}{dx}=9x^2-1$

a) Now this is $\displaystyle 8$ when:

$\displaystyle 9x^2-1=8$,

or:

$\displaystyle x^2=1$

or $\displaystyle x=\pm 1$, which may be plugged into the equation for y to
give $\displaystyle y=10$ when $\displaystyle x=1$ and $\displaystyle y=6$ when $\displaystyle x=-1$.
So the required points are $\displaystyle (1,10)$ and $\displaystyle (-1,6)$.

b)Now this is $\displaystyle 0$ when:

$\displaystyle 9x^2-1=0$,

or:

$\displaystyle x^2=1/9$

or $\displaystyle x=\pm 1/3$.

Plugging these values back into the equation for $\displaystyle y$ gives the other
coordinate, and so we find the required points are: $\displaystyle (1/3,8)$ and $\displaystyle (-1/3,8)$.

RonL

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.
The points at which the curve crosses the t-axis correspond to the roots of:

$\displaystyle 4t^2+5t=0$

which are $\displaystyle t=0$ and $\displaystyle t=-5/4$.

Now:

$\displaystyle \frac{ds}{dt}=8t+5$,

so the slopes at the two points are $\displaystyle 5$ and $\displaystyle -5$ respectively.

Now the gradient of the normal is -1 over the gradient of the tangent, and so
the gradients of the two normals are $\displaystyle -1/5$ and $\displaystyle 1/5$ respectively.

RonL

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)
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.

RonL

10. 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.
Sorry the question should be:

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)