VCE Methods Question Thread!

[One Step at a Time]

Ah got it, thanks so much RuiAce  You always reply so quickly eheh

[RuiAce]

4. (Also 3rd pic) For 8h) , the answer for the implied domain= (-∞,2) ∪ [ 1,∞], but I only got [1,∞]. My logic (probs not logical lol) is that (x-1)/(x-2)≥0 and x+2≠0. So x≥1 and x≠-2, which means that the implied domain is [1,∞]. 

2. (2nd pic) For Q6), can we work out the range without drawing the graph? I resorted to drawing the graph for each equation to figure out the range, but I'm not sure if it's the quickest/ most effective way to do so.

Unless you can visualise it in your head you probably have to draw it.

3. (3rd pic) For 8c), does the fact that the discriminant < 0/ there are no X-int for x^2+3≥0 affect the implied domain? Or is it totally irrelevant to calculate the implied domain?

[clarke54321]

Could someone please give me some guidance with this question. Thanks!

[RuiAce]

Could someone please give me some guidance with this question. Thanks!

[vcestressed]

Hi everyone,
Why are these polynomials 'non-factorable' into a perfect square?

x^2+14x-49
4x^2-10x+25
x^2+2/3x-1/9

i know that for the first one (x-7)^2 doesn't add up to x^2+14x-49. . . but how can you know this /without/ working it out fully?

[rosalie.brown]

Hey clarke54321,

I've included the solutions to your question below, despite using a very similar method to RuiAce, hehe.

Find the equations of the lines that pass through the point (1,7) and touch the parabola:

1. Using the point-slope formula, substitute the given point (1,7):
y-y1 = m(x-x1) --> Let y1 = 7, x1 = 1.
Hence y-7 = m(x-1), gives equation 1: y=mx-m+7

2. Let equation 2 => y=-3x^2+5x+2 (i.e. the inital parabola). Now let equation 1 and equation 2 equal one another to give:
mx+7-m = -3x^2+5x+2
3x^2-5x-2+mx+7-m=0
3x^2+(m-5)x+ (5-m)=0

3. The equation 3x^2+(m-5)x+(5-m)=0 is the new quadratic (mentioned in the hint section) that is required to solve the problem. This quadratic is still in the form y=ax^2+bx+c. Let the discriminant of this equation equal 0.

0=b^2-4ac, whereby b=(m-5), a=3, c=(5-m)
0=(m-5)^2-4(3)*(5-m)
0=(m^2-10m+25)-12(5-m)
0=m^2+2m-35
0=(m-7)(m-5)
∴ m = -7 or m = 5

4. Let m = -7. Substitute this value into y=mx+c, along with the point (1,7):
y=-7x+c
7=-7(1) + c
∴ c = 14
∴ y=-7x+14

5. Let m=5. Substitute this value into y=mx+c, along with the point (1,7):
y=5x+c
7=5(1) + c
∴ c = 2
∴ y=5x+2

∴ Your solutions are y=-7x+14 and y=5x+2.
---
Hopefully you found this post helpful!

[clarke54321]

Thanks for the help RuiAce and rosalie.brown! 

[RuiAce]

Hi everyone,
Why are these polynomials 'non-factorable' into a perfect square?

x^2+14x-49
4x^2-10x+25
x^2+2/3x-1/9

i know that for the first one (x-7)^2 doesn't add up to x^2+14x-49. . . but how can you know this /without/ working it out fully?

_____________

[RuiAce]

Hey clarke54321,

I've included the solutions to your question below, despite using a very similar method to RuiAce, hehe.

Find the equations of the lines that pass through the point (1,7) and touch the parabola:

1. Using the point-slope formula, substitute the given point (1,7):
y-y1 = m(x-x1) --> Let y1 = 7, x1 = 1.
Hence y-7 = m(x-1), gives equation 1: y=mx-m+7

2. Let equation 2 => y=-3x^2+5x+2 (i.e. the inital parabola). Now let equation 1 and equation 2 equal one another to give:
mx+7-m = -3x^2+5x+2
3x^2-5x-2+mx+7-m=0
3x^2+(m-5)x+ (5-m)=0

3. The equation 3x^2+(m-5)x+(5-m)=0 is the new quadratic (mentioned in the hint section) that is required to solve the problem. This quadratic is still in the form y=ax^2+bx+c. Let the discriminant of this equation equal 0.

0=b^2-4ac, whereby b=(m-5), a=3, c=(5-m)
0=(m-5)^2-4(3)*(5-m)
0=(m^2-10m+25)-12(5-m)
0=m^2+2m-35
0=(m-7)(m-5)
∴ m = -7 or m = 5

4. Let m = -7. Substitute this value into y=mx+c, along with the point (1,7):
y=-7x+c
7=-7(1) + c
∴ c = 14
∴ y=-7x+14

5. Let m=5. Substitute this value into y=mx+c, along with the point (1,7):
y=5x+c
7=5(1) + c
∴ c = 2
∴ y=5x+2

∴ Your solutions are y=-7x+14 and y=5x+2.
---
Hopefully you found this post helpful!

Improvement: If you used y-y₁=m(x-x₁) then you don't have to use y=mx+b

Look back in 1. You already know that y=mx-m+7. Just sub straight back into there.

[Izzy999]

Hey guys,
I've been staring at this question for a couple of days now and I'm not really understanding the order of how I should do things.

1. For each of the following, find (f+g)(x) and (fg)(x) and state the domain for both f+g and fg:

b) f(x)=1-x^2 for all x ∈ [-2,2] and g(x)=x^2 for all x ∈ R+

Any help would be appreciated, thanks

[peanut]

Hey guys,
I've been staring at this question for a couple of days now and I'm not really understanding the order of how I should do things.

1. For each of the following, find (f+g)(x) and (fg)(x) and state the domain for both f+g and fg:

b) f(x)=1-x^2 for all x ∈ [-2,2] and g(x)=x^2 for all x ∈ R+

Any help would be appreciated, thanks

(f+g)(x) = 1 - x^2 - x^2
= 1 - 2x^2
(fg)(x) = x^2(1-x^2)
= x^2 - x^4

In general, domain for (fg)(x) and (f+g)(x) = dom f   n   dom g

So, for both, domain = (0,2]

[Syndicate]

(f+g)(x) = 1 - x^2 - x^2
= 1 - 2x^2
(fg)(x) = x^2(1-x^2)
= x^2 - x^4

In general, domain for (fg)(x) and (f+g)(x) = dom f   n   dom g

So, for both, domain = (0,2]

(f+g)(x) = 1

Everything else seems correct 

[peanut]

(f+g)(x) = 1

Everything else seems correct 

Oh right, careless mistake. Thanks!

[deStudent]

I'm having trouble solving simultaneous literal equations with my calc. It says "too few arguments..."

The problem was to solve for x and y in:

3(x - a) - 2(y + a) = 5 - 4a
2(x + a) + 3(y - a) = 4a - 1

I solved it by hand and got: x = a + 1 and y = a - 1

But in general I'm having trouble solving simultaneous literal equations with the cas calc..

[Syndicate]

I'm having trouble solving simultaneous literal equations with my calc. It says "too few arguments..."

The problem was to solve for x and y in:

3(x - a) - 2(y + a) = 5 - 4a
2(x + a) + 3(y - a) = 4a - 1

I solved it by hand and got: x = a + 1 and y = a - 1

But in general I'm having trouble solving simultaneous literal equations with the cas calc..

Menu->3:Algebra->7:Solve System of Equations->2: Solve System of Linear Equations

Write the equations in the blank spaces, and press enter.