Question
Evaluate slope of a tangent line:
y=1-(1/2)x^2, P(2,-1)
The answer is supposed to be -1, but I'm getting -2.
[1-2-2h-(1/2)h^2]-[-1]/h
[-1-2h-(1/2)h^2+1]/h
[h(-2-1/2h)]/h
-1/2(0)-2
=-2
y=1-(1/2)x^2, P(2,-1)
The answer is supposed to be -1, but I'm getting -2.
[1-2-2h-(1/2)h^2]-[-1]/h
[-1-2h-(1/2)h^2+1]/h
[h(-2-1/2h)]/h
-1/2(0)-2
=-2
Answers
Damon
first using calculus to check
y=1-(1/2)x^2, P(2,-1)
dy/dx = -x
at x = 2
That is
-2
Like you are right, period.
y=1-(1/2)x^2, P(2,-1)
dy/dx = -x
at x = 2
That is
-2
Like you are right, period.
bobpursley
y=1-(1/2)x^2
y'=0-2/2 x
slope= -x=-2
The lim method:
=1-(1/2)(x+h)^2 -1+1/2(x)^2
= 1-1/2)(x^2+2xh+h^2)-1+1/2(x)^2 all over h
as h > 0
= ( -xh-h^2)/h= -x at at 2,-1
=-2
y'=0-2/2 x
slope= -x=-2
The lim method:
=1-(1/2)(x+h)^2 -1+1/2(x)^2
= 1-1/2)(x^2+2xh+h^2)-1+1/2(x)^2 all over h
as h > 0
= ( -xh-h^2)/h= -x at at 2,-1
=-2