With transformations of [3,2] and [2,-2], i am to find the image of the following equations(noting that each of the above transformations are treated separately for each equation:

a)y=|x|

b)y=x^2-3x

Printable View

- Feb 18th 2011, 05:44 PMjohnsy123Matrix Transformations
With transformations of [3,2] and [2,-2], i am to find the image of the following equations(noting that each of the above transformations are treated separately for each equation:

a)y=|x|

b)y=x^2-3x - Feb 18th 2011, 05:48 PMmr fantastic
- Feb 18th 2011, 06:16 PMjohnsy123
i am critically most stuck on applying the transformations(they're are horizontal and vertical) of [3,2] and [2,-2] to the equation x^2-3x.

i factorized x^2-3x which came out to be x(x-3) and then i applied the transformations seperately, which is what is asked, but then it got a lil messy and wrong. - Feb 18th 2011, 06:20 PMProve It
Here's how to do part b) i), you follow the same procedure to do part b) ii)...

If you remember back to when you studied quadratics, the quadratic equation $\displaystyle \displaystyle y = a(x - h) + k$ represents a translation of $\displaystyle \displaystyle (h, k)$ to the equation $\displaystyle \displaystyle y = ax^2$.

So for your quadratic $\displaystyle \displaystyle y = x^2 - 3x$

$\displaystyle \displaystyle = x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2$

$\displaystyle \displaystyle = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$,

applying the transformation $\displaystyle \displaystyle (3, 2)$ gives

$\displaystyle \displaystyle y = \left(x - \frac{3}{2} - 3\right)^2 - \frac{9}{4} + 2$

$\displaystyle \displaystyle = \left(x - \frac{9}{2}\right)^2 - \frac{1}{4}$

$\displaystyle \displaystyle = x^2 - 9x + \frac{81}{4} - \frac{1}{4}$

$\displaystyle \displaystyle = x^2 - 9x + 20$.