# Math Help - Questions on Graph transformations

1. ## Questions on Graph transformations

Hello! I am a little confused about graph transformations and I have some questions that I don't really know how to solve...Please help! Thank you!

1)
The transformations A,B and C are as follows:
A: A translation of 1 unit in the negative y direction
B: A stretching parallel to the y-axis (with x-axis invariant) with a scale factor of 2
C: A reflection about the x-axis

A curve undergoes in succession, the above transformations A,B and C and the equation of the resulting curve is y = -2(x^2) -6. Determine the equation of the curve before the transformations were effected. [Is it possible to view the equation of the resulting curve as y= -2 (x^2 -3)? But I think this will give a different original equation from if I use the given resulting equation??]

2)
The equation of a graph is y = (ax)/(bx-y), and y= c is the horizontal asymptote while x = d is the vertical asymtote of this graph. Given that c = d = 1/2, what are the values of a and b? [If I am not wrong, b should be 2, but I don't know how to find a....Is a = 1?]

3)
Consider the curve (5x + 10)^2 - (y-3)^2 = 25. Prove, using an algebraic method, that x cannot lie between two certain values (which are to be determined). [I know how the given equation can be expressed to become the standard equation of a hyperbola, but I am not sure what is the "algebraic method" the question is asking for...]

Thank you very very much for helping me!

2. Originally Posted by Tangera
...
3)
Consider the curve (5x + 10)^2 - (y-3)^2 = 25. Prove, using an algebraic method, that x cannot lie between two certain values (which are to be determined). [I know how the given equation can be expressed to become the standard equation of a hyperbola, but I am not sure what is the "algebraic method" the question is asking for...]
From

$(5x+10)^2-(y-3)^2=25~\iff~25(x+2)^2-(y-3)^2=25~\iff~$ $\frac{(x+2)^2}{1^2} - \frac{(y-3)^2}{5^2} = 1$

you know:

The center of the hyperbola is C(-2, 3)

The vertices of the hyperbola are $V_1(-3, 3)$ , $V_2(-1,3)$

Therefore $x \notin (-3, -1)$

Second attempt

Solve the equation for x:

$x = -2\pm\frac15 \cdot \sqrt{25+(y-3)^2}$

The smallest value the radical can be is 25 if $(y-3) = 0$

The value of x lies between:

$-2-\frac15 \cdot 5 < x < -2+\frac15 \cdot 5$

$-3 < x < -1$

3. Originally Posted by earboth

Second attempt

Solve the equation for x:

$x = -2\pm\frac15 \cdot \sqrt{25+(y-3)^2}$

The smallest value the radical can be is 25 if $(y-3) = 0$

The value of x lies between:

$-2-\frac15 \cdot 5 < x < -2+\frac15 \cdot 5$

$-3 < x < -1$
Ermm...sorry could you explain why do you want the value in the radical to be smallest? Thank you!