How do you sketch the graph of #y=-4(x-3)^2+2# and describe the transformation?

1 Answer
Feb 5, 2018

See answer below.

Explanation:

If you are studying transformations and want to understand how to easily see all the effects for any function, I recommend memorising this general function. It's really worthwhile if you have exams on this stuff.

#f(x)=a(n(x-b))^k+c#

#a# is the dilation factor from the x-axis. If #a# is negative, the function is reflected in the x-axis.

#1/n# is the dilation factor from the y-axis. If #1/n# is negative, the function is reflected in the y-axis.

#b# is the horizontal translation. If #b# is positive, the function is shifted to the right, if it is negative, the function is shifted to the left.

#c# is the vertical translation. If it is positive, the function is shifted upwards, if it is negative, the function is shifted downwards.

In this case, the parent function is #y=x^2#

and it can have up to four transformations applied to it to get

#y=-4(x-3)^2+2#

Now, lets find the four parameters and describe the transformations from the parent function.

#a=-4#

The parent function is dilated by a factor of 4 from the x-axis and reflected in the x-axis.

#1/n=1#

There is no dilation effect from the y-axis.

#b=3#

The function is shifted 3 units to the right.

#c=2#

The function is shifted 2 units up.

To graph, notice that the function is already in turning point form so you can read it straight off as #(3,2)#. The parabola is upside down because #a# is negative, so now you just need to find the x and y-intercepts.

y-intercept (set #x=0#):

#y=-4(0-3)^2+2=-34#

x-intercepts (set #y=0#):

#0=-4(x-3)^2+2rArr1/2=(x-3)^2rArrx=3+-sqrt(1/2)#
graph{-4(x-3)^2+2 [-10, 10, -5, 5]}