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anonymous
 4 years ago
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 xaxis?
anonymous
 4 years ago
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 xaxis?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have no idea how to do this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=x^2+x+13\]\[f'(x)=2x+1\] Any line is parallel to the xaxis 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
 4 years ago
Best ResponseYou've already chosen the best response.0@above? how is the line vertical and tangent to that graph at the same time?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i see, maybe, sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The line is horizontal, because the xaxis is horizontal. Two horizontal lines are parallel.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i didnt see that it said xvalue, i put the actual line that was tangent and parallel

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and i messed that part up too!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the equation for the tangent line would be y = 51/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The equation for the tangent line would be: \[y=\frac{1}{2}x+13\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no, plug the axis of symmetry (1/2) in for x to get 12 3/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and yours isnt even horizontal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0y = 13? That would touch it at two points though. You got the vertex, now you just need a horizontal line with the ycoordinate of the vertex so that it touches the vertex.
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