Thread: Functions help (test soon)

Hey guys, i have a test soon and just wanted to get a few things cleared up. Thanks

1. a. How do u identify the x-intercepts of the graph of y=x(x^2-9a^2) (note: a>0)

b. How would u go about sketching the graph of y= x(x^2-9a^2)

2. If the equation 2x^2 + 6=ax has exactly one solution, find the value(s) of a, such that this is true.

3. Is it possible to find the x intercepts of the function f(x)= 4x^3 - 3x^2- 8x + 5 without using a calculator?

4. If the graph of f: R arrow R crosses the x axis exactly three times, which one of the following rules could not be the rule for f?
A. y=x(x^2-4)
B. y=x(x-2)(x+4)(x^2+1)
C. y=(3-x)(x^4-16)
D y=(x^2-x-6)(x-4)
E y=(x^2-x-12)

5. For the curve of the function with equation y= (2-x)(x-2)(x-5), the subset of R for which the gradient of the graph is positive is best described by?

Originally Posted by chaneliman

1. a. How do u identify the x-intercepts of the graph of y=x(x^2-9a^2) (note: a>0)

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

x intercepts are

x+0 = 0 , x = 0
x-3a=0 , x= 3a
x+3a =0 , x= -3a

Originally Posted by chaneliman

b. How would u go about sketching the graph of y=
x(x^2-9a^2)

power of 3 we start with a positive gradient … unless there was a reflection in the x

from left to right pass though -3a 1st then loop back 0 then loop back around to 3a …

Originally Posted by chaneliman

2. If the equation 2x^2 + 6=ax has exactly one solution, find the value(s) of a, such that this is true.

2x^2 –ax + 6 = 0

then take the Discriminate and set it to 0

b^2 - 4ac = 0

a^2 - 4(12)= 0

a^2 = 48

a= +root48 or –root48

Originally Posted by chaneliman

3. Is it possible to find the x intercepts of the function f(x)= 4x^3 - 3x^2- 8x + 5 without using a calculator?

yes … but you will be here all day … and need to know polynomial division which would take a while to teach.. best thing to do .. solve for y=0

Originally Posted by chaneliman

4. If the graph of f: R arrow R crosses the x axis exactly three times, which one of the following rules could not be the rule for f?
A. y=x(x^2-4)
B. y=x(x-2)(x+4)(x^2+1)
C. y=(3-x)(x^4-16)
D y=(x^2-x-6)(x-4)
E y=(x^2-x-12)

E because you can only see power of 2… if you expand it so max amount of crosses is 2 …

Originally Posted by chaneliman

5. For the curve of the function with equation y= (2-x)(x-2)(x-5), the subset of R for which the gradient of the graph is positive is best described by?

ok rearrange y= (-x+2)(x-2)(x-5)

y= -(x-2)(x-2)(x-5)
y= -(x-2)^2(x-5)

Intercepts are … 2 and 5 and 2 is a turning point

ok because its power of the graph is 3 we start positive but we have reflection in x so we actually start negative… we get to 2 bounce up doesn’t not cross remember then we start to get a positive gradient sadly… it returns to pass though 5 … so we need to differentiate the graph…

y= (-x+2)(x-2)(x-5) = -x^3 + 9x2 -24x+20

dy/dx = -3x^2+18x-24

let that equal to 0 to find stationary points

-3x^2+18x-24=0

x^2-6x+8=0

(x-2)(x-4)=0

therefore 2 and 4 … and that’s where the graph is positive … 2 < x < 4

thanks for doing all that. just one question wot do u mean by "because you can only see power of 2… if you expand it so max amount of crosses is 2 …"

Originally Posted by chaneliman

thanks for doing all that. just one question wot do u mean by "because you can only see power of 2… if you expand it so max amount of crosses is 2 …"

yer sorry didnt really proof read ... i ment expand the brackets and look at the highest power.. in this case it was x^2 and power of 2 means it crosses the x axis 2 times...

to cross three times you needed at least a power of three e.g x^3

can someone help me find the equation of the following

Originally Posted by chaneliman

can someone help me find the equation of the following

I have to make a few assumptions:

1. The function is at least a cubic function.
2. (2, 2) is a turning point.That means f(2) = 2 and f''(2) = 0
3. At (2, 2) the graph has a horizontal tangent. That means f'(2) = 0
4. The graph passes through the origin. That means f(0) = 0

If so:

Use the genral equation of a cubic function:







From #4 you know d = 0

Plug in the coordinates (2, 2) into the equations and solve the following system of simultaneous equations for (a, b, c):

. After a few steps you should get

Originally Posted by chaneliman

can someone help me find the equation of the following

Assuming that the function is a cubic and assuming (2, 2) is a stationary point of inflexion (earboth, I'm sure that's the word you meant rather than turning point), the following model can be immediately used:

.

To get a, the assumption that (0, 0) is point on the curve will do the trick:

.

Then .

This answer is equivalent to the one found by earboth.

By the way, the point (4, 4) also appears to lie on the curve - our answers are certainly consistent with that.

thanks. Could u clear some things up for me?

1. why does the equation x^2-3x move to the right by 3? (i mean its not a translation)

2. how would u describe the tanformations of the graph y=3x^3 + 5x + x +1? (like dilations, transations, reflections)

2. has 4/x been dilated by the y-axis (not the x-axis, i know how to do it from the x axis) by a factor of 4 or 1/4?

Originally Posted by chaneliman

thanks. Could u clear some things up for me?

1. why does the equation x^2-3x move to the right by 3? (i mean its not a translation) Mr F says: Move to the right relative to what graph?? The question makes no sense.

2. how would u describe the tanformations of the graph y=3x^3 + 5x + x +1? (like dilations, transations, reflections)

Mr F says: Transformations compared to what graph?? Again, the question makes no sense.

2. has 4/x been dilated by the y-axis (not the x-axis, i know how to do it from the x axis) by a factor of 4 or 1/4?

Mr F says: Dilation from the y-axis: f(x) --> f(ax) and the dilation factor is 1/a.

f(x) = 1/x --> f(x/4) = 4/x. a = 1/4 and so the dilation factor is 4.

..

1. relative to x^2

2. relative to x^3

Originally Posted by chaneliman

1. relative to x^2

Mr F says: Complete the square: .

So .

You should be able to easily read off the transformations involved.


2. relative to x^3

Mr F says: Well then, bad luck sport. You see, the thing is that has turning points (at x = -1 and x = -1/9). But has a stationary point of inflexion. It's therefore impossible to get from using any combination of translations, dilations or reflection in coordinate axes. So that's that.

..

1. The largest domain which the function f(x) = 1/square root x-4 is defined by, is: [4, ~) or (4,~ ) and how did u noe?

2. For the curve of the function with the equation y=(2-x)(x-2)(x-5), the subset of R for whcih the gradient of the graph is positive is best described by?

3. The parabola with the equation y=x^2 is translated so that its image has a vertex (a,b). The equation of the image is:

4. The linear factors of x^2-9x are: I said it was 0, -3, 3 but the books says thats wrong and the answer in x, x-3, x+3

5. 3 points have coordinates A (1,7), B (7,5), C(0, -2). Find the equation of the line perpendicular to AB, passing through the midpoint odf BC?

6. Given that f(x+4) = x^3-4, then f(x) is? i said it was (x+4)^3-4 but the book says its (x-4)^3-4 y is that?