VCE Methods Question Thread!

[IndefatigableLover]

Thanks
Just forgot that intersection of y is when x=0 -_-"

[Phy124]

Find the equation of the tangent to the curve with the equation
y=3x³-4x²+2x-10
at the point of intersection with the 'y' axis.

 

Firstly find derivative;



Sub in x=0 as this is where the graph will intersect with the y-axis



Therefore the derivative at this point is 2.

Secondly, find the y-coordinate at this point such that you can find the equation of the line.

Let x = 0 in the original equation



So at the point (0,-10) you have a gradient of 2.

To find the equation of this line, use;









edit: beaten and you know what to do now, never mind

I should probably pay attention to that red text and see what people have typed before posting 

[IndefatigableLover]

Another quick question:

The curve with the equation
y=(x-2)(x-3)(x-4) cuts the 'x' axis at the points
P(2,0) , Q(3,0) , and R(4,0)

At what point does the normal to the curve at Q cut the 'Y' axis?

[Phy124]

Another quick question:

The curve with the equation
y=(x-2)(x-3)(x-4) cuts the 'x' axis at the points
P(2,0) , Q(3,0) , and R(4,0)

At what point does the normal to the curve at Q cut the 'Y' axis?

1. Differentiate the equation
2. Find the gradient at Q (x=3)
3. Sub x=3, y = 0 and (Where a is the gradient at Q, as the normal is the negative reciprocal of the gradient of the curve) into
4. Sub x=0 into the equation you get from 3 to give you the y-intercept of line

edit: Forgot to mention, due to the symmetry of the graph, the gradient through Q will just be -1, but you won't really get questions like these, so best to get into a habit of differentiating.

[IndefatigableLover]

Thanks

[ecvkcuf]

Lol year 9 and already doing Year 12 methods? wtf..

my self-esteem just plummeted

[Hutchoo]

Lol year 9 and already doing Year 12 methods? wtf..

my self-esteem just plummeted

And he/she likes Yonghwa -- automatic pro right there.

[IndefatigableLover]

And he/she likes Yonghwa -- automatic pro right there.

LOL questioning my gender like that
And I see you're a fan of Taeyeon

Also another question...

The function s(t)= -3t^3+6t^2-3 represents the displacement of a particle moving along a straight line, where 't' is in seconds and 's' is in metres.

a). Find the position of the particle after 3 seconds.
b). Find the velocity of the particle at that time.

[ligands]

might be wrong here but i think

a) sub t = 3 into s(t), s(3) = -3*3^3+6*3^2-3 = 6, therefore the position is 6 to right of the origin
b) sub t=3 into s'(t), velocity at t=3 is s'(x) = -9t^2 + 12t, s'(3) = -9*3^2 + 12*3 = -45, therefore the velocity at t=3 and position=6 is -45 showing that the particle is decreasing

someone correct this if its wrong

[IndefatigableLover]

might be wrong here but i think

a) sub t = 3 into s(t)
b) sub t=3 into s'(t)
someone correct this if its wrong

.... Damn got to read the question properly when I do a question next time -.-

I also have one last question:

Consider the equation y=x(x^2-9)

a). Find the gradient at the points at which the curve crosses the x-axis .
b). Find the coordinates of the point on the curve at which the gradient =0

[TrueTears]

a) x(x^2-9)=0

x(x+3)(x-3) = 0

x = 0, -3, 3

sub in those numbers into dy/dx

b) solve dy/dx = 0, sub those x values into y(x)

[IndefatigableLover]

a) x(x^2-9)=0

x(x+3)(x-3) = 0

x = 0, -3, 3

sub in those numbers into dy/dx

b) solve dy/dx = 0, sub those x values into y(x)

Thank you

[ligands]

was i correct henry?

a) from graphing y=x(x^2-9), we find that the x ints are at x=3, x=0, x=-3
therefore y=x^3 - 9x, dy/dx=3x^2 - 9
subbing the x ints into dy/dx to find the gradient at the x intercepts

b) set dy/dx = 0, therefore 3x^2 - 9 = 0, solve for x we find x = -sqroot(3) and x = sqroot(3)
sub these 2 x values into y=x^3 - 9x to find the y value at gradient = 0

sorry if its messy i dont know how to do the neat pictures like others yet
someone correct if something is wrong

[IndefatigableLover]

Yeah it's correct
Just forgot that 't=time in seconds' -_-"

And thanks again for working out the other question

[AutumnConcerto]

I have a question:

Is y=3(x-3)^2 considered a perfect square or not? :S