## At what point is the tangent to f(x) = x^(2) + 4x -1 horizontal?

Differential equationsAt what point is the tangent to f(x) = x^(2) + 4x -1 horizontal? 9 months ago
At what point is the tangent to f(x) = x^(2) + 4x -1 horizontal?
For the tangent to be horizontal, the slope of the graph at that point has to be zero. We find the slope function of f: lim_(h-> 0)([(x+h)^(2)+4(x+h)-1]-(x^(2)+4x-1))/(h)=lim_(x-> 0)((x+h)^(2)+4(x+h)-1-x^(2)-4x+1)/(h) =lim_(h-> 0)((x+h)^(2)+4(x+h)-x^(2)-4x)/(h) =lim_(h-> 0)(x^(2)+2hx+h^(2)+4x+4h-x^(2)-4x)/(h) =lim_(h-> 0)(2hx+h^(2)+4h)/(h) =lim_(h-> 0)[2x+h+4] =2x+4 Then if the slope is zero: 2x+4=0-> x=-2 And so the point of intersection is (-2, f(-2))=(-2,-5) Result: (-2,-5)
Use definition for derivative.
lim_(h-> 0)(f(a+h)-f(a))/(h)
Plug (a+h) into problem.
=([(a+h)^(2)+4(a+h)-1]-(a^(2)+4a-1))/(h)
Simplify and combine like terms.
=([a^(2)+2ah+h^(2)+4a+4h-1]-(a^(2)-4a-1))/(h)
factor out h.
=(2ah+h^(2)+4h)/(h)=(h(2a+h+4))/(h)
Solve limit by pluggin in 0.
lim_(h-> 0)2a+h+4=2a+(0)+4=2a+4
Horizontal tangent has slope of 0, to set a=0 and solve by subtracting 4 and dividing by 2.
2a+4=0
2a=-4
a=-2
Plug -2 back into original function to find y value.
(-2)^(2)+4(-2)-1=-5
Result:
(-2,-5)
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