Variety of Maths Help Please - Urgent :(

[hello_kitty]

Hey Guys,
Ive been doing heaps of revision for my sac on monday was just wondering if you could help me out with a few questions and how to work them out.
The ones in red aren't urgent!

Gallery of Graphs

1) The equation of the circle that has a diameter with end points at (4,-5) and (-4,-5) is what?
I'm not sure how to work out these types of questions.. any tips?
Answer: x^2 + (y+5)^2 = 16

2) The graph of y = 1/(cx-a) + b, where a and b are positive constants, has axis intercepts of what?
Answer (ab-1)/bc

3) For the graph of y = 1/(x-3)^2  - 4 ... Why is 'The graph is symmetric about the y-axis' not true? and why is 'the y-coordinate of the graph are all greater than -5' true??

4) Let (c,d) be a point on the circle with equation x^2 + y^2 = a^2 (c does not equal a, AND c does not equal 0)
a) Find d in terms of a and c
b) Find the equation of the tangent to the circle at (c,d) in terms of c and d
c) Find the y axis intercept of the tangent in terms of a and c
d) Find the x axis intercept of the tangent in terms of a and c

5a) Find the coordinate of the points of interesection of the straight line with the equation y=x-a and the circle with equation x^2 + y^2 = 4. Give the coordinates on a for: 2 points of intersection         and      the line to be tangent to the circle


Functions and Relations

1) The function with rule f(x) = mx + 2 , m < 0, has an inverse function with rule f^-1(x) = ax + b, a,b the element of R.
Why is: a < 0, b > 0 true??

2) For f: (a,b] --> R, f(x) = 5-x, where a <b, the range is?   Why is the range [5-b, 5-a)


Also, with Truncas' for example y = 7/(x+4)^2 Is the maximal domain R\{-4} and to work out the range it would be the horizontal asymptote to infinity? In this case (0, infinity)?

Whereas, with Hyperbolas, to figure out the range and domain it would just be R\{horizontal asymptote) and R\{vertical asymptote}?

Any helpw ould be awesome! thank yoU!!!

[xZero]

1) the diameter of a circle is the difference in the extreme position of x or y value, in this case d=4-(-4)=8, hence radius is 4. The centre of the circle is the mid point of the extreme positions, hence mid point = (0,-5), now construct the equation of the circle (x-a)^2+(y-b)^2=r^2, where the centre of the circle is (a,b). Plug in the numbers and you'll get x^2+(y+5)^2=16

2) x-int, y=0. 0=1/(cx-a)+b, -b=1/(cx-a), cx-a=1/-b, cx=1/-b +a, x = 1/-bc + a/c, x = (ab-1)/bc

3)The graph is not symmetric about the y-axis is because its been shifted to the right by 3 units, so it should be symmetric about x=3 instead.
The minimum y value of y = 1/(x-3)^2  - 4 is approaching -4. Hence the y co-ordinates will be greater than -5 (if your confused just graph it on a calculator)

functions
1) y=mx+2, lets find the inverse of this equation. x=my+2, x-2=my, y=x/m - 2/m which equates to f^-1(x)=ax+b. Thus a=1/m and b=-2/m. We know that m<0, then a<0 and b>0.

2) f(x)=5-x, this graph has a negative gradient so the lower limit of the domain should have a higher y-value. if x=a, the f(a)=5-a. If x=b, then f(b)=5-b. Remember that the lower limit will have a greater y-value, hence the range is [5-b,5-a)

[aznxD]

Q1 - The general equation for a circle is (x-h)^2 + (y-k)^2 = r^2
Where (h,k) is the centre of the circle and r is the radius.
since the diameter has end points at (4,-5) and (-4,-5) the centre will be the midpoint.
((4+-4) / 2 , (-5+-5) /2)
= (0,-5)
The diameter is 8 units therefore the r=4.
Sub them into the general equation and you get x^2 + (y+5)^2 = 16

As for Domain and range it is easier to sketch the graph. Once you sketch the graph the domain (x-values) and range (y-values) become much more clearer.

[PlayerKay]

2) For f: (a,b] --> R, f(x) = 5-x, where a <b, the range is?   Why is the range [5-b, 5-a)
Since the domain of the function is (a,b] and since the equation is a negative linear, the lowest point on the graph is when x is at its highest value. The highest point on the graph will be when x is at its lowest value.
The range can be found simply by subbing in the values for x given in the domain.
Lowest value for y is when x is at its highest value. Lowest value for x=b.
Highest value for y is when x is at its lowest value x approaches a but does not equal it, indicated by the round bracket.
Subbing those 2 values in gives the range [5-b , 5-a)

About the Truncas question and hyperbola, in general that is the correct way to figure out domain and range. But remember if a Truncas is flipped along the x-axis it is to -∞.

[RobM8]

Q4. a) sub in the point (c,d): (c)^2 + (d)^2 = a^2
     rearrange to get: d = ±√(a^2-c^2)
     b) differentiate with respect to x: 2x + 2y(dy/dx) = 0
     rearrange to get: dy/dx = -x/y
     sub in the point (c,d): dy/dx = m = -c/d, hence tangent = (-c/d)x + C (= mx + c)
     sub in (c,d) to your tangent equation: d = (-c/d)(c) + C
     rearrange: C = d + c^2/d = (d^2 + c^2)/d
     equation of tangent: y = (-c/d)x + (d^2 + c^2)/d
     c) set x=0: y = (-c/d)(0) + d + c^2/d
     clean it up: y = (d^2 + c^2)/d
     sub in d in terms of a and c from part a): y = ±((a^2 - c^2) + c^2)/√(a^2 + c^2) = ±a^2/√(a^2 + c^2)
     state intercepts: (0,√(a^2 + c^2)) or (0,-√(a^2 + c^2))
     d)set y=0, do the same thing: (0) = (-c/d)x + (d^2 + c^2)/d
     in terms of x: x = (d/c)((d^2 + c^2)/d) = d(d^2 + c^2)/(cd) = (d^2 + c^2)/c
     sub in d in terms of a and c from part a): x = ((a^2 - c^2) + c^2)/c = a^2/c
     state intercept: (a^2/c,0)
Q5. a) sub the linear equation into the circle: x^2 + (x-a)^2 = 4
     expand and collect like terms: x^2 + x^2 - 2ax + a^2 = 4, therefore 2x^2 - 2ax + a^2 = 4
     solve for x using the general quadratic formula: (cant be fucked typing it out) x = .5(a ±√( 8 - a^2))
     sub our x value into one of the equations: y = .5(a ±√( 8 - a^2)) - a = .5(±√( 8 - a^2) - a)
     state the points of intersection: (.5(a + √( 8 - a^2)),.5(√( 8 - a^2) - a)) or (.5(a - √( 8 - a^2)),-.5(√( 8 - a^2) + a))

[hello_kitty]

thanks all for your help!

[MissIraq]

i have question

if the angle between the lines 2y=8x+10 and 3x-6y+22 is j, then tan J is..? jesus someone help me.

[RobM8]

use tan(theta)=m
where m is the gradient of the line and theta is the angle between the line and the positive x axis.

3x-6y+22 is an expression, there needs to be an equals sign somewhere in it.

[MissIraq]

idk, hey what about..
f(x)=e^(-x) and g:(-1,infit)-->R where g(x)= loge(x+2) the fuction with the rule y=f(g(x) has a rangE? ..?

[RobM8]

Probs should start your own question thread instead of hi-jacking someone else's.

Composite Function f(g(x)) --> ran g has to be a subset or equal to dom f for the composite function to exist.

ran g = [0,∞) and dom f = R

ran g is a subset of dom f and hence the composite function exists.

dom f(g(x)) = dom g = (-1,∞)

f(g(x)) = e^-(loge(x+2)) = e^(loge((x+2)^-1))) = 1/(x+2)

ran y = ran 1/(x+2) over the domain (-1,∞)

f(g(-1)) = 1

As x --> ∞, f(g(x)) --> 0 from above

therefore ran y = (0,1)