# calculus

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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 -

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 -

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?

• calculus -

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 -

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 -

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 -

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 -

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 -

how on earth did you get x=-.568

the equation is right you just solved it wrong.

• calculus -

I'm sorry all, I apparently can't do basic math.

Thank you everyone for helping me.
I really appreciated it.

• calculus -

x=-7.1

• calculus -

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 -

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)

y=2(-7.1)+14=-.2 (or -1/5)

• calculus -

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 -

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 -

>Reiny

spoon fed the answer right there, so yeah never plug it in to the equation of the line.

• calculus -

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.

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