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 x-axis?

1. anonymous

i have no idea how to do this

2. anonymous

$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

3. anonymous

y = -2

4. anonymous

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

5. anonymous

oh i see, maybe, sorry

6. anonymous

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

7. anonymous

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

8. anonymous

and i messed that part up too!

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

and yours isnt even horizontal

13. anonymous

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

14. 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.