Author Topic: VCE Methods Question Thread! (Read 5712702 times) Tweet Share

ETTH96 · « **Reply #3840 on:** January 23, 2014, 05:38:02 pm »

How do I find the asymptote of an exponential function?

And how would I find the turning point?

f(x)=xe^-x

For turning point, would I differentiate and let it equal 0?

Anchy · « **Reply #3841 on:** January 23, 2014, 06:50:04 pm »

Having some issues with this question:

The parabola has TP at (z,-8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5; Find the value of z.

b^3 · « **Reply #3842 on:** January 23, 2014, 07:12:42 pm »

If we have a parabola with turning point at $(h,k)$ then we have a parabola of the form $y=a(x-h)^{2}+k$ . You can substitute in the known turning point. We then have two unknowns, $a$ and $z$ . But we have two pieces of information, $(0,10)$ and $(5,0)$ , which you can substitute into the equation and solve simultaneously for $z$ . Hope that helps

EDIT:

Spoiler

$\begin{alignedat}{1}y & =a\left(x-z\right)^{2}-8 \\ \left(0,10\right) \\ 10 & =a\left(0-z\right)^{2}-8 \\ az^{2} & =18\qquad[1] \\ \left(5,0\right) \\ 0 & =a\left(5-z\right)^{2}-8 \\ a & =\frac{8}{\left(5-z\right)^{2}}\qquad[2] \\ \text{Substitute [2] into [1]} \\ \frac{8}{\left(5-z\right)^{2}}z^{2} & =18 \\ 8z^{2} & =18\left(5-z\right)^{2} \\ 4z^{2} & =9\left(25-10z+z^{2}\right) \\ 5z^{2}-90z+225 & =0 \\ z^{2}-18z+45 & =0 \\ \left(z-15\right)\left(z-3\right) & =0 \\ \therefore z & =3,\:15 \end{alignedat}$

You don't need to find $a$ in this case, but this is what it comes out as if you needed to.
$\begin{alignedat}{1}z=3 \\ a & =\frac{8}{4} \\ & =2 \\ z=15 \\ a & =\frac{8}{100} \\ & =\frac{2}{25} \end{alignedat}$

If you want to picture it graphically, then this is what we have:
https://www.desmos.com/calculator/pp4guicx7k

M_BONG · « **Reply #3843 on:** January 24, 2014, 02:08:04 am »

I really need to clear up the basics before I screw up on harder questions..
Can anyone explain what I have done wrong please?

solve for x: $4-3x^{2}\leq 0$
$4\leq 3x^{2}$
$x^{2}\geq \frac{4}{3}$
$x\geq \pm \frac{2}{3^\frac{1}{2}}$

Actual answer: $\frac{2}{3^{\frac{1}{2}}}\leq x\leq \frac{-2}{3^{\frac{1}{2}}}$

Thanks!

Phy124 · « **Reply #3844 on:** January 24, 2014, 02:28:28 am »

Quote from: Zezima. on January 24, 2014, 02:08:04 am

I really need to clear up the basics before I screw up on harder questions..
Can anyone explain what I have done wrong please?

solve for x: $4-3x^{2}\leq 0$
$4\leq 3x^{2}$
$x^{2}\geq \frac{4}{3}$
$x\geq \pm \frac{2}{3^\frac{1}{2}}$

Actual answer: $\frac{2}{3^{\frac{1}{2}}}\leq x\leq \frac{-2}{3^{\frac{1}{2}}}$
Thanks!

$4-3x^{2}\leq 0$

$4\leq 3x^{2}$

$x^{2}\geq \frac{4}{3}$

$|x| \geq \frac{2}{\sqrt{3}}$

$x \geq \frac{2}{\sqrt{3}}$ and $-x \geq \frac{2}{\sqrt{3}}$

$x \geq \frac{2}{\sqrt{3}}$ and $x \leq -\frac{2}{\sqrt{3}}$

$\frac{2}{\sqrt{3}} \leq x \leq -\frac{2}{\sqrt{3}}$

If you're dealing with inequalities of quadratic functions it can often be helpful to sketch the scenario

lzxnl · « **Reply #3845 on:** January 24, 2014, 02:32:37 am »

x^2 >= 4/3 means squaring x is greater than 4/3.
First solve for the equality sign. We get x=+-sqrt(4/3)
The thing about inequalities is that the solution must be in a region(s) between where equality holds, or outside this region.
Well...if x is in between -sqrt(4/3) and +sqrt(4/3), squaring x gives you something less than 4/3, right? So that region doesn't work. We're left with x>=sqrt(4/3) and x<=-sqrt(4/3)

Anchy · « **Reply #3846 on:** January 24, 2014, 01:48:56 pm »

Quote from: b^3 on January 23, 2014, 07:12:42 pm

If we have a parabola with turning point at $(h,k)$ then we have a parabola of the form $y=a(x-h)^{2}+k$ . You can substitute in the known turning point. We then have two unknowns, $a$ and $z$ . But we have two pieces of information, $(0,10)$ and $(5,0)$ , which you can substitute into the equation and solve simultaneously for $z$ . Hope that helps

EDIT:
Spoiler
$\begin{alignedat}{1}y & =a\left(x-z\right)^{2}-8 \\ \left(0,10\right) \\ 10 & =a\left(0-z\right)^{2}-8 \\ az^{2} & =18\qquad[1] \\ \left(5,0\right) \\ 0 & =a\left(5-z\right)^{2}-8 \\ a & =\frac{8}{\left(5-z\right)^{2}}\qquad[2] \\ \text{Substitute [2] into [1]} \\ \frac{8}{\left(5-z\right)^{2}}z^{2} & =18 \\ 8z^{2} & =18\left(5-z\right)^{2} \\ 4z^{2} & =9\left(25-10z+z^{2}\right) \\ 5z^{2}-90z+225 & =0 \\ z^{2}-18z+45 & =0 \\ \left(z-15\right)\left(z-3\right) & =0 \\ \therefore z & =3,\:15 \end{alignedat}$

You don't need to find $a$ in this case, but this is what it comes out as if you needed to.
$\begin{alignedat}{1}z=3 \\ a & =\frac{8}{4} \\ & =2 \\ z=15 \\ a & =\frac{8}{100} \\ & =\frac{2}{25} \end{alignedat}$

If you want to picture it graphically, then this is what we have:
https://www.desmos.com/calculator/pp4guicx7k

Thank you so much!

M_BONG · « **Reply #3847 on:** January 24, 2014, 01:58:32 pm »

Quote from: Phy124 on January 24, 2014, 02:28:28 am

$x^{2}\geq \frac{4}{3}$

$|x| \geq \frac{2}{\sqrt{3}}$

Thanks!

But I don't understand why you took the absolute function of 'x'. Ie. what logic you used to get from the 1st line to the 2nd?

Kuroyuki · « **Reply #3848 on:** January 24, 2014, 02:02:40 pm »

Quote from: Zezima. on January 24, 2014, 01:58:32 pm

Thanks!

But I don't understand why you took the absolute function of 'x'. Ie. what logic you used to get from the 1st line to the 2nd?

I think its like
sqrt (x^2) is the same as mod x

psyxwar · « **Reply #3849 on:** January 24, 2014, 03:23:12 pm »

Quote from: Zezima. on January 24, 2014, 01:58:32 pm

Thanks!

But I don't understand why you took the absolute function of 'x'. Ie. what logic you used to get from the 1st line to the 2nd?

x^2=4
x=+2, -2

If we had to work out what x^2>4 was, then:

The modulus of x values greater than +2 would be acceptable (eg. |3| = 3, which when squared yields 9). The modulus of x-values less than 2 would also be acceptable (eg. |-3|=3, which when squared also gives 9). It isn't true to say that x^2>4, and therefore x>+2 and x>-2, because you can see that this isn't true (eg. |-1|=1, 1^2=1 which is less than 4). Therefore we say that the modulus of x has to be greater than that value, rather than the value of x itself.

HEN_iP · « **Reply #3850 on:** January 25, 2014, 04:35:06 pm »

If the functions
$f(x)=x^4+ax^3+bx^2+36x+144$
and
$g(x)=x^4+(a+3)x^3-22x^2+(b+10)x+40$
both cross the $x$ axis at -2, determine the values for both $a$ and $b$

I have no idea how to approach this.

Phy124 · « **Reply #3851 on:** January 25, 2014, 04:58:11 pm »

Quote from: HEN_iP on January 25, 2014, 04:35:06 pm

If the functions
$f(x)=x^4+ax^3+bx^2+36x+144$
and
$g(x)=x^4+(a+3)x^3-22x^2+(b+10)x+40$
both cross the $x$ axis at -2, determine the values for both $a$ and $b$

I have no idea how to approach this.

$f(-2)=(-2)^4+(-2)^3a+(-2)^2b+36(-2)+144 = 0$

$\Rightarrow 2a - b = 22 \ ... (1)$

$g(-2) = (-2)^4 + (-2)^3(a+3) - 22(-2)^2 + (b+10)(-2) + 40 = 0$

$\Rightarrow 4a + b = -38 \ ... (2)$

Solve them simultaneously and you should get $a=-\frac{8}{3}$ and $b = -\frac{82}{3}$

Jason12 · « **Reply #3852 on:** January 25, 2014, 05:42:49 pm »

how do you derive an expression that is a fraction? E.g.

$\dfrac{dy}{dx}\dfrac{4x^3}{x^3}$

Stick · « **Reply #3853 on:** January 25, 2014, 05:45:26 pm »

Quote from: Jason12 on January 25, 2014, 05:42:49 pm

how do you derive an expression that is a fraction? E.g.

$\dfrac{dy}{dx}\dfrac{4x^3}{x^3}$

Do you mean $\frac{d}{dx}(\frac{4x^3}{x^3})$ ?

The expression simplifies to $\frac{d}{dx}(4)$ , which is just $0$ .

alchemy · « **Reply #3854 on:** January 25, 2014, 06:00:24 pm »

Quote from: Jason12 on January 25, 2014, 05:42:49 pm

how do you derive an expression that is a fraction?

Use the power rule.
Best way to go about this is to always simplify the expression first.
Example: f(x) = 5/x
We can simply this to: 5*x^-1
Now we can derive as normal to get:
5(-1)*x^-2=-5*x^-2
Put this back into fractional form to get:
-5/x^2

Sorry for the lack of LaTeX, I'm kinda in a rush. Hope that helped though. Have a read of the power rule if you're interested in seeing how this works with various cases.

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