calculus
posted by Natalie .
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

calculus 
WaywardSon
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. 
calculus 
Natalie
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? 
calculus 
MathMate
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. 
calculus 
WaywardSon
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. 
calculus 
Natalie
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. 
calculus 
Natalie
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. 
calculus 
RePost
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. 
calculus 
WaywardSon
how on earth did you get x=.568
the equation is right you just solved it wrong. 
calculus 
Natalie
Okay. My bad.
I'm sorry all, I apparently can't do basic math.
Thank you everyone for helping me.
I really appreciated it. 
calculus 
WaywardSon
x=7.1

calculus 
Natalie
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. 
calculus 
WaywardSon
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) 
calculus 
Reiny
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 
calculus 
WaywardSon
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. 
calculus 
RePost
>Reiny
spoon fed the answer right there, so yeah never plug it in to the equation of the line. 
calculus 
Natalie
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.
Respond to this Question
Similar Questions

calculus
Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent point to the xaxis. If the tangent point is close to the yaxis, the line segment is long. If the tangent … 
12th Grade Calculus
1. a.) Find an equation for the line perpendicular to the tangent curve y=x^3  9x + 5 at the point (3,5) [* for a. the answer that I obtained was y5 = 1/18 (x3) ] b.) What is the smallest slope on the curve? 
Calculus
Consider the curve y^2+xy+x^2=15. What is dy/dx? 
calculus
1. Given the curve a. Find an expression for the slope of the curve at any point (x, y) on the curve. b. Write an equation for the line tangent to the curve at the point (2, 1) c. Find the coordinates of all other points on this curve … 
calculus
Consider the curve defined by 2y^3+6X^2(y) 12x^2 +6y=1 . a. Show that dy/dx= (4x2xy)/(x^2+y^2+1) b. Write an equation of each horizontal tangent line to the curve. c. The line through the origin with slope 1 is tangent to the curve … 
calculus
A curve passes through the point (1,11) and it's gradient at any point is ax^2 + b, where a and b are constants. The tangent to the curve at the point (2,16) is parallel to the xaxis. Find i) the values of a and b ii) the equation … 
Calculus 1
The curve y =x/(sqrt(5−x^2)) is called a bulletnose curve. Find an equation of the tangent line to this curve at the point (2,2). 
Calculus 1
The curve y = x/(sqrt(5 x^2)) is called a bulletnose curve. Find an equation of the tangent line to this curve at the point (2, 2) 
Calculus AB
Sorry but I've got a lot of problems that I don't understand. 1) Let f(x)= (3x1)e^x. For which value of x is the slope of the tangent line to f positive? 
Calculus AB
Could someone please help me with these tangent line problems?