Write an equation of the translated or rotated graph in general form.
y=3x^2-2x+5 T(2,-3)
I know the graph is a parabola. But I am not sure where I would "plug in" h and k into this equation. To get the translated equation.
1. You know
$\displaystyle \left|\begin{array}{l}\bar{x} = x+2 \\ \bar{y} =y-3\end{array}\right. ~\implies~ \left|\begin{array}{l}\bar{x}-2 = x \\ \bar{y}+3 =y\end{array}\right.$
2. Plug in the terms for x and y and change the variable:
$\displaystyle y+3 = 3(x-2)^2-2(x-2)+5$
Expand and collect like terms.
I am having difficulty with one more question
y^2+8x=0 rotated on pi/6
I did this
x= cos(pi/6)x+sin(pi/6)y
square root 3 / 2 x+1/2y
For y I got -1/2X+ square root 3/2y
Then I plug it in
(-1/2x+ square root 3/2y)^2+8( square root 3 /2 x +1/2y)
But I am having trouble figuring out the equation.
1. Do yourself (and us) a favour and start a new thread if you have a new question.
2. I assume that the center of rotation is the origin(?). If so you know:
$\displaystyle \left|\begin{array}{l}\bar{x} = x \cdot \cos\left(\frac{\pi}6\right) - y \cdot \sin\left(\frac{\pi}6\right) \\ \bar{y} = x \cdot \sin\left(\frac{\pi}6\right) + y \cdot \cos\left(\frac{\pi}6\right) \end{array}\right.$
With $\displaystyle \cos\left(\frac{\pi}6\right)=\frac12 \sqrt{3}$ and $\displaystyle \sin\left(\frac{\pi}6\right) = \frac12$
you'll get:
$\displaystyle \left|\begin{array}{l}x = \frac12 ( \bar{x} \sqrt{3} + \bar{y}) \\ y= \frac12 (\bar{y} \sqrt{3} - \bar{x} )\end{array}\right.$
3. Plug in these terms into the original equation which yields:
$\displaystyle \frac14 (\bar{y} \sqrt{3} - \bar{x})^2+ 4( \bar{x} \sqrt{3} + \bar{y}) = 0$
4. Expand the brackets and collect like terms:
$\displaystyle x^2 - 2·√3·x·y + 16·√3·x + 3·y^2 + 16·y = 0$
I have a question when you wrote the equation why is it square root of 3 instead of square root of 3 over 2? I mean it makes sense in the final answer.
Also would instead of having 1/2 in the equation would you not have 2 because if you plug it back into the equation with two it makes more sense to me
As I checked and found out the answer to this problem is
x^2-2square root 3 xy +3(y)^2+ 16square root 3 x+16y
And if you use 2 instead of 1/2 in the equation I would get that answer.
You probably have noticed that I made a mistake typing the equation (I used √3 instead of \sqrt{3}) So Here is the correct equation:
$\displaystyle x^2 - 2\cdot \sqrt{3} \cdot x·y + 16\cdot \sqrt{3} \cdot x + 3·y^2 + 16·y = 0$
Sorry for the confusion.