calculus
posted by Natalie on .
Find the point on the curve y=1 x2+14 x such that the tangent line to the curve is parallel to 2 x + 10 y = 23

2x+10y=23 (is what im guessing since the problem breaks awkwardly there)
y=1/5x+23/10 (slope intercept form)
derivative of y=x^2+14x (im guessing since it looks like you expressed it wrong)
y'=2x+14
I'll leave it up to you at that point. 
Yes. Sorry.
I did get it up to the derivative of y' = 2x + 14 by using the
f'(x) = [f(x+h)  f(x)] / h
formula.
and did simplify the other equation into the slopeintercept form.
but do I just plug the derivative into the 'slopeintercept' equation?
because isn't y' = 2x+14 the slope of the tangent.
I guess I'm mostly confused because I need to find a point parallel to that second equation and for the point to be parallel, they must have the same slope.
yet the derivative gave me a different slope? 
For slope intercept form,
y=mx+b
m is the slope.
You need to find the value of x for which
f'(x)=m, i.e. tangent parallel to y=mx+b. 
A slope can't be an equation. It's a number, and your solving for x so that the equation equals the slope of the line.
Also your teacher taught you the long way of doing a derivative, if the "Power Rule" for derivatives never comes up in class you're better off looking it up yourself. 
I'm sorry. For the life of me, I just can't figure out this problem.
Okay, so if the slope can't be an equation, can i solve for x by setting y' equal to zero?
or do you mean
2x + 14 = (1/5) x + (23/10)
I tried both ways I just listed, and they turned out wrong, so I'm still lost.
He did teach us the shortcut during my last class. It was just a force of habit to do it the long way.
Thank you so much for helping me. 
I've also just tried:
1/5 = 2x + 14
x = .568
and plugged that into the 'parallel' equation to solve for y.
y= 2.4136
but that still didn't work out. 
what's the slope of the line?
set 2x+14 equal to that and solve.
(hint:slope intercept form y=mx+b)
sorry if i sound like a twat but trust me that you'll remember things better if you figure them out yourself. 
how on earth did you get x=.568
the equation is right you just solved it wrong. 
Okay. My bad.
I'm sorry all, I apparently can't do basic math.
Thank you everyone for helping me.
I really appreciated it. 
x=7.1

Alright. One last question.
So I'm wondering if my online homework answer is wrong, because I have x and I've plugged x back into the y = (1/5)x + (23/10)
and I've solved for y.
y = 3.72
Now please tell me if I'm just making a fool of myself again and am incorrectly solving an algebraic equation or is it wrong, because it won't accept 3.72 as the y, but accepts 7.1 as the x. 
you don't plug it into the equation of the line.
what you are trying to do is to get the derivative equal to the slope of the line y=1/5x+23/10
y=mx+b
so the slope, m=(1/5)
therefore 1/5=2x+14
where x=(7.1)
check your solution
y=2(7.1)+14=.2 (or 1/5) 
my oh my oh my !!!
why did you not follow MathMate's idea?
y' = 2x + 14, that is your slope at any point (x,y)
the slope of the given straight line is 1/5
so 2x + 14 = 1/5
x = 7.1 You had that!
then y = (7.1)^2 + 14(7.1) = 149.81
so the point of contact is (7.1, 149.81)
since the new line is parallel to the old, it must have the same x and y terms, so
2x + 10y = c
plug in the point
2(7.1) + 10(149.81) = c = 1512.3
equation:
2x + 10y = 1512.3 
derivative is a slope. so y'=..... is not the same as y=.....
think of y' as m, a slope as well but tangent to graph. 
>Reiny
spoon fed the answer right there, so yeah never plug it in to the equation of the line. 
Wow thank you guys all so much.
I'm dreading a whole semester of calculus.
(stats was so much easier)
Thanks especially for explaining everything, so now I understand it.