Let f(x)=x^2+x+13. What is the value of x for which the tangent line to the graph of y=f(x) is parallel to the x-axis?

Question

anonymous · Answer

i have no idea how to do this

xishem · Answer

$$f(x)=x^2+x+13$$$$f'(x)=2x+1$$

Any line is parallel to the x-axis when its slope is 0, so find x for when the tangent line's slope is equal to 0:

0 = 2x+1
-1=2x
x = -1/2

anonymous · Answer

y = -2

anonymous · Answer

@above? how is the line vertical and tangent to that graph at the same time?

anonymous · Answer

oh i see, maybe, sorry

xishem · Answer

The line is horizontal, because the x-axis is horizontal. Two horizontal lines are parallel.

anonymous · Answer

yes i didnt see that it said x-value, i put the actual line that was tangent and parallel

anonymous · Answer

and i messed that part up too!

anonymous · Answer

the equation for the tangent line would be y = 51/4

xishem · Answer

The equation for the tangent line would be:
$$y=\frac{-1}{2}x+13$$

anonymous · Answer

no, plug the axis of symmetry (-1/2) in for x to get 12 3/4

anonymous · Answer

and yours isnt even horizontal

xishem · Answer

I apologize. The equation for the tangent line would be: $$y=13$$
$$(\frac{-1}{2},\frac{51}{4}) $$is actually the vertex point for the original equation

anonymous · Answer

y = 13? That would touch it at two points though. You got the vertex, now you just need a horizontal line with the y-coordinate of the vertex so that it touches the vertex.