How do you sketch the graph of #y=(x+4)^2-5# and describe the transformation?

1 Answer
Jul 19, 2017

Move the graph of #x^2# four units to the left and five units down.

Explanation:

Okay, so this quadratic equation is called the turning point form. Notice how that if you expand the expression in the brackets, there will be no coefficient for x, thus, the graph will not dilate or become narrow.

This is called the turning point form because the instructions on how to move the graph are in the equation itself.

The constant outside the bracket represents the number of units you will have the move on the y-axis. In this case, it is #-5#, so you will move it five units down.

But wait! Wouldn't you move the graph four units to the right because the constant in the brackets is positive? Yes, you would. However, the sign of the constant in the brackets must be change to the opposite. In this case, the positive sign will change into the negative sign.

All you have to do is move the graph of #x^2# four units to the left and five units down.

All you have left to do is to find the y and x-intercept. The y-intercept can be found by just turning the #x# into a zero. Once you do that, solve for y, which is 11.

#y=(x+4)^2-5#
#y=(0+4)^2-5#
#y=16-5#
#y=11#

The x-intercept can be found by using the null factor law. Since the #5# is negative, we can use the difference of squares law to put the five into the brackets. To do this, look below:

#y=(x+4)^2-5#
#y=(x+4-sqrt5)(x+4+sqrt5)#
#y=(x+1.76)(x+6.24)#
Apply null factor law.
#x+1.76=0#
#x=-1.76#
or
#x+6.24#
#x=-6.24#
Those are the two x-intercepts, all you have to do is to plug in these coordinates and connect the dots!

Hope this helps!