2 Answers

Best Answer

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)

Best Answer

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)

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)