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Calc math homework help, please.

H

Howard

Lifer
Find the equation of the tangent of y=x^2 + 6x + 9 with a slope of 4.

I know it's y = 4x + b, I just need to find b.
 
WTF do your own homework

j/k

slope = dy/dx

then you just need the point of tangency which you can solve by solving dy/dx = 4, then subbing x into original euqstion for y

then sub x,y into your line eq and solve for b
 
no i use aim sorry

dy/dx does give you enough to work with

dy/dx = 2x + 6 = 4, easily solve for x
then sub x into original equation for y
 
Originally posted by: dighn
no i use aim sorry

dy/dx does give you enough to work with

dy/dx = 2x + 6 = 4, easily solve for x
then sub x into original equation for y
Click to expand...
How did you get 2x + 6?
 
Howard, are you starting afresh? Do you know how to take the deriv of the original equation y=x^2 + 6x + 9 ? The deriv comes from there.

If you need to start with the Derivative by definition (4 step), see your book or see here. Basically, you're taking the delta y over delta x when the deltas are extremely small, hence the added on h approaching zero.

If you want the shortcut: you're basically taking the x^2, multiplying the whole thing by a coefficient of the power (2) and dropping the power by 1 from 2 to 1. On the other term, you have 6x, so you multiply by the power of 1 then reduce the x^1 power to x^0, leaving your product as 6.
 
do you know the limits definition of derivative? you can use that

otherwise you'd have to do this graphically or something.
 
plug -1 into both equations, find the value of the original function at x= -1 and make b whatever value will make the tangent function equal to it
 
If you only know limits, then you can either get ahead in class and learn how to do derivatives or you can draw out the graph and find the equation of the line tangent to y which gives you m = 4. Been too long since I've done limit problems. 🙂
 
well, if all you know is limits, you may be able to solve it with the defination of a derivative (aka derivatives the long way). This is usually taught before telling "students what a derivative is so students dont take the derivative for granted" (what my calc 1 prof said). Anyways, the equation is (f(x+h)-f(x))/h. i think that is the formula. However, i dont know how you can get a limit juts by knowing teh slope if you have no concept of the derivative.
 
Bah. I used to think solving derivatives by first principles using limits was the most annoying thing in calc class.
Ever since going to college and doing a senior year math class, i've since learnt that first principles are frequently more important than knowing the derivatives themselves.

Simply because it gets so complex you can't use the fixed formulas after a while...
 
Originally posted by: Ionizer86
Howard, are you starting afresh? Do you know how to take the deriv of the original equation y=x^2 + 6x + 9 ? The deriv comes from there.

If you need to start with the Derivative by definition (4 step), see your book or see here. Basically, you're taking the delta y over delta x when the deltas are extremely small, hence the added on h approaching zero.

If you want the shortcut: you're basically taking the x^2, multiplying the whole thing by a coefficient of the power (2) and dropping the power by 1 from 2 to 1. On the other term, you have 6x, so you multiply by the power of 1 then reduce the x^1 power to x^0, leaving your product as 6.
Click to expand...
Well, I can find the tangent of any point on the graph of x^2 + 6x + 9, but my problem is I can't find the points that correspond to a slope of 4 on that graph.
 
Originally posted by: dighn
WTF do your own homework

j/k

slope = dy/dx

then you just need the point of tangency which you can solve by solving dy/dx = 4, then subbing x into original euqstion for y

then sub x,y into your line eq and solve for b
Click to expand...

 
Originally posted by: Ionizer86
If you want the shortcut: you're basically taking the x^2, multiplying the whole thing by a coefficient of the power (2) and dropping the power by 1 from 2 to 1. On the other term, you have 6x, so you multiply by the power of 1 then reduce the x^1 power to x^0, leaving your product as 6.
Click to expand...
😕

 
Well, I can find the tangent of any point on the graph of x^2 + 6x + 9, but my problem is I can't find the points that correspond to a slope of 4 on that graph.
Click to expand...

That' what a derivative's for. Gives you the instantaneous dy/dx (slope) at any point on that line.

Well, I can find the tangent of any point on the graph of x^2 + 6x + 9, but my problem is I can't find the points that correspond to a slope of 4 on that graph.
Click to expand...

Well, it's kinda "cheating" in a way using the shortcut. See your textbook for the definition of a derivative: lim (h-->0) [f(x+h)-f(x)]/h

It's basically taking your current point, going up an extremely small increment h (next-to-zero sized increment), then finding out the difference between the x+h point and the x point. Hence, since the increment is really really small, we consider it the instantaneous slope.

Once again, your textbook will be your best guide (unless you feel like reading the web site), or you can wait till next time to have the teacher/prof explain this stuff.
 
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