What is the equation of the line that is normal to #f(x)= (x+2)^2-2x^2+2 # at # x=-1 #?
2 Answers
# y = -1/6x+5/6 #
Explanation:
The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is
We have:
# f(x) = (x+2)^2-2x^2+2 #
# " " = x^2+4x+4-2x^2+2 #
# " " = -x^2+4x+6 #
Differentiating wrt
# dy/(dx) = -2x+4 #
So when
# y = -1-4+6=1# and#dy/dx=2+4=6#
So the tangent/normal passes through
# y - 1 = -1/6(x+1) #
# y - 1 = -1/6x-1/6 #
# y = -1/6x+5/6 #
We can verify this graphically:
Explanation:
One way to think of the derivative is as a "slope-finding-formula". The result of a derivative is a function, such that when you plug in values for
The normal line is just the perpendicular line through the tangent line, at the same point. Recall from algebra that if you have a slope
Perpendicular slope:
First, determine the exact point where your normal line will pass through the given curve. This happens at
The point that our normal line will go through is
Second, find the derivative, or slope-finding formula. It may help to fully expand out the original function first.
Then find the derivative
The slope of the tangent line at
Therefore, the slope of the perpendicular normal line is
Now you have all the parts you need to determine the formula for this normal line. You have a slope,