Posted by Natalie on Tuesday, September 28, 2010 at 11:11pm.
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 slope-intercept form.
but do I just plug the derivative into the 'slope-intercept' 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.