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?

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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?

Mathematics
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i have no idea how to do this
\[f(x)=x^2+x+13\]\[f'(x)=2x+1\] 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
y = -2

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@above? how is the line vertical and tangent to that graph at the same time?
oh i see, maybe, sorry
The line is horizontal, because the x-axis is horizontal. Two horizontal lines are parallel.
yes i didnt see that it said x-value, i put the actual line that was tangent and parallel
and i messed that part up too!
the equation for the tangent line would be y = 51/4
The equation for the tangent line would be: \[y=\frac{-1}{2}x+13\]
no, plug the axis of symmetry (-1/2) in for x to get 12 3/4
and yours isnt even horizontal
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
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.

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