Question

Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. (If an answer does not exist, enter DNE.)
y = 1/2 x^2 + 7x

Answer 1

When a line is horizontal, what is its slope?

Answer 2

Should be 0, correct? ;)

Answer 3

So then if you take the derivative of (1/2)x^2 + 7x and set it equal to 0, what equation do you have?

Answer 4

So you would get (2)(1/2)x^1 + 7 = 0 or x + 7 = 0, wouldn't you?

Answer 5

it'sasking for an x,y coordinate

Answer 6

Yes, I know :)

Answer 7

You want to find the point, or x,y coordinates of the point, where the slope of the tangent line is 0 (horizontal).

Answer 8

So if you take the derivative of the function and set it equal to 0, you can figure out what the x-coordinate of the point must be. Since the derivative is x + 7, if we set x + 7 = 0 and solve for x

Answer 9

what must x be?

Answer 10

If x + 7 = 0, I mean

Answer 11

If x + 7 = 0, then x must be -7 ...

Answer 12

And when x = -7, if you go an plug that value in for x back in your original function, you will get the y-coordinate that goes with that x.

Answer 13

So, y = 1/2(-7)^2 + 7(-7)

Answer 14

24.5

Answer 15

Please let me know what you get when you do that calculation and I will work it as well. :)

Answer 16

You're faster than I am. Just a moment, please.

Answer 17

Yes, I got -24.5.

Answer 18

So y = -24.5
So then your point where the tangent line is horizontal is (x,y) = (-7, -24.5)

Answer 19

Which does make sense if you think about it. This is a quadratic equation, so the graph is a parabola. and so the tangent line will be horizontal at the vertex, which is (-7,-24). So that makes sense. :)

Answer 20

Thank you Anteater :)

Answer 21

You are very welcome! :)