Determine when y = x^2 + bx + 17 has horizontal tangent

[gillgill]

am i doing this right?
Determine a value of the constant b so that the graph of y=x^2+bx+17 has a horizontal tangent at (2, 21+2b)
ans: y'(2)=0
y'=2x+b
0=2(2)+b
b=-4
what do u need the y-coordinate (21+2b)for?
My teacher also requires me to draw a diagram...can anybody show me a diagram of this problem?

 

[Ouabache]

Did you try plotting your original equation with the value you calculated for b?
From that plot you will see why you need to use the point(2,21+2b).
(Hint: it has to do with the graphical interpretation for taking the first derivative of your equation)

If you are having a problem drawing the diagram.. Try calculating some ordered pairs using a range of x values (say from -10 to 10) and plug that into your original equation. You already have b, so you should be able to calculate all your y-values. Then plot these ordered pairs on some graph paper.

 

[gillgill]

my friend taught me that
21+2(-4)=13
(2)^2+(-4)(2)+17=13
i am not too sure why he did that...

 

[Ouabache]

my friend taught me that
21+2(-4)=13
(2)^2+(-4)(2)+17=13
i am not too sure why he did that...

What does that suggest to you about the point(2,13) and the two
equations you are now working with?

After you plot the ordered pairs as i suggested, from your initial equation y(x)
Do the same thing for the second equation y'(x) and note what is happening.

 

[hypermorphism]

am i doing this right?
...

You could have also done it by finding the correct vertex of the parabola.

...
what do u need the y-coordinate (21+2b)for?

Nothing.

My teacher also requires me to draw a diagram...can anybody show me a diagram of this problem?

It's a standard parabola translated a little. Complete the square to find the x and y offsets.

 

[Ouabache]

How did you make out plotting your curves?
Did you see why need to use the point(2,21+2b)?