anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i have no idea how to do this
Xishem
  • Xishem
\[f(x)=x^2+x+13\]\[f'(x)=2x+1\] 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
  • anonymous
y = -2

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