Given f(x) = 2 x^2 + 4 x + 5 , determine the point on the curve where the tangent line to the curve is horizontal. (The answer must be a POINT on the curve.)
The point on the curve where the tangent line is horizontal is (). (Separate the coordinate values with a comma.)

Question

Given f(x) = 2 x^2 + 4 x + 5 , determine the point on the curve where the tangent line to the curve is horizontal. (The answer must be a POINT on the curve.)
The point on the curve where the tangent line is horizontal is (). (Separate the coordinate values with a comma.)

anonymous · Answer

Differentiate, set the derivative to zero, solve for x.  Substitute that value into f to get the appropriate y value.  Show answer as (x,y).

anonymous · Answer

thanks, this was helpful

anonymous · Answer

(x,Y) = (-1,3)

anonymous · Answer

$$f(x) = 2 x^2 + 4 x + 5$$$$f'(x)=4x+4=0\rightarrow x=-1\rightarrow f(-1)=3$$So the solution is (x.y)=(-1,3)

anonymous · Answer

I apparently don't keep up on typing skills.....

anonymous · Answer

lol

anonymous · Answer

By the way, you could use algebra to find the x coordinate of the vertex (x=-b/2a) to get the same result with less sophistication.

anonymous · Answer

cool, good to know for the future :)

anonymous · Answer

Of course, that only works on quadratics; the calculus method works in general.

anonymous · Answer

ok