cherylim23's methods question thread :)

[the.watchman]

That is because, as indicated by the eqn, the graph has a stationary point of inflexion at (a,0), so the graph would keep decreasing until another stationary point, definitely between a and b

[|ll|lll|]

Another question (which I kinda not get)
Find the value of k for which a - 3b is a factor of a^4 - (7a^2)(b^2) + k(b^4).
Hence, for this value of k, factorise a^4-7a^2b^2 + kb^4 completely.

k = -18 (fyi)

Btw, I need to learn latex

[brightsky]

In order to factor this, we must try to make have factors and/or





Let





Now, it is apparent that in order to factorise THIS, we need

Because

We need

[|ll|lll|]

Oh okay thanks!
I see, but aren't we supposed to use factor theorem for this question...?

[brightsky]

Oh! Yes! Much easier that way!

Substitute in into the expression for all the . Because it is a factor, by the factor theorem, it would equal to 0.

[Yitzi_K]

The factor theorem states that if (x-a) is a factor of f(x), then f(a)=0. So to use the factor theorem you just sub in the a value.

[selim31]

@cherylim23
Hey, how did you go on your SAC? I had mine today and it was totally easy

[99.95]

how do i do this question:

Let P and Q be points on tge curve y=x^2 at which x=1 and x= 1+ h respectively. Express the gradient of the line PQ in terms of h and hence find the gradient of the tangent to the curve y=x^2 at x=1

[Yitzi_K]

how do i do this question:

Let P and Q be points on tge curve y=x^2 at which x=1 and x= 1+ h respectively. Express the gradient of the line PQ in terms of h and hence find the gradient of the tangent to the curve y=x^2 at x=1

Gradient = y₂ - y₁ over x₂ - x₁.

This gives you (f(x+h) - f(x))/h

Simplify, (down to 2x+h) then let h=0 (even though h can't equal 0, we let it tend towards zero, which comes to the same thing) then sub in x=1.

This is called the First Principle method.

[brightsky]

Ok, the point P is for a y-coordinate and point Q is for a y-coordinate .

The points are on the curve hence it satisfies the equation.

Substitute in the coordinates:





Hence the points are P and Q .

To find the gradient of the line, use the formula: .

So Gradient =

As for your second question, fastest way is the differentiate



Substitute x = 1,

Gradient =

[99.95]

how do i do this question:

Let P and Q be points on tge curve y=x^2 at which x=1 and x= 1+ h respectively. Express the gradient of the line PQ in terms of h and hence find the gradient of the tangent to the curve y=x^2 at x=1

thanks for the help

Gradient = y₂ - y₁ over x₂ - x₁.

This gives you (f(x+h) - f(x))/h

Simplify, (down to 2x+h) then let h=0 (even though h can't equal 0, we let it tend towards zero, which comes to the same thing) then sub in x=1.

This is called the First Principle method.

Ok, the point P is for a y-coordinate and point Q is for a y-coordinate .

The points are on the curve hence it satisfies the equation.

Substitute in the coordinates:





Hence the points are P and Q .

To find the gradient of the line, use the formula: .

So Gradient =

As for your second question, fastest way is the differentiate



Substitute x = 1,

Gradient =

thanks for the help

[brightsky]

Yitzi_K's method is essentially the same. In full: Differentiation by first principles. Differentation comes from that. The derivative is essentially the gradient.

[m@tty]

As for your second question, fastest way is the differentiate



Substitute x = 1,

Gradient =

The question says "hence", so you must use your answer to the previous part of the question.

You must let h=0 so .

You let h=0 because for the chord PQ to be a tangent it must have only one point of intersection with the parabola, ie. P=Q or h=0.

But you can certainly use brightsky's method to check your answer.

[99.95]

how do i differentiate:

y=(e^x^2)^5?

[superflya]

same as e^5x^2. now differentiate normally. it should be 10xe^5x^2