Find an equation for the line tangent to the graph of the given function at the indicated point.
f(x) = x2-3 at (3,6)

- anonymous

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- schrodinger

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- anonymous

The first thing you need to do is find out how fast the function f(x) is changing at the given point. And to do that you need the derivative f '(x). Would you like help finding the derivative?

- anonymous

Yes, please

- anonymous

You can look at the parts of f(x) individually.
Umm, is the function f(x) = 2x - 3 or f(x) = x^2 - 3?

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- anonymous

\[f(x)= x ^{2}-3 at (3,6)\]

- anonymous

Nice.
So to find the derivative of the x^2 part, you need to use the power rule. The two comes down and becomes the coefficient, and that is raised to 2-1 = 1. Which is just x so I'm going to leave it out.
The derivative of the - 3 is zero, because the derivative is the rate of change and the rate of change of anything constant is 0.
The whole derivative is the sum of the parts:
f ' (x) = 2x + 0
Which is just f ' (x) = 2x.

- anonymous

Next you need to find out what the derivative is when x = 3.
f ' (3) = 2 * 3
f ' (3) = 6
So that is going to be the slope m in your y = mx + b for the tangent.
Hopefully this makes sense, I've never tutored anyone before in my life. ;D

- anonymous

It's telling me the answer should be y=2x-12

- anonymous

Thank you, your actually helping me see things I haven't noticed. I'm sorry, this course is really hard for me. I've been doing my homework for like 4 hours :(

- anonymous

No prob. I wish I got the *$(#) problem right though. ;/

- anonymous

No, it shouldn't be that line y = 2x - 12. I just plotted them both on my graphing calculator and they don't even intersect.

- anonymous

And I plotted y = 6x - 12 and it looks right.
Should we go on working through the problem and see if we get - 12 for the y intercept?

- anonymous

It is 6x-12, I'm sorry I must not being seeing right. I'm getting tired of learning lol. How do you plot it and get an equation?

- anonymous

Once you have figured out the slope from the derivative, it is really easy to get the equation for the tangent line.
You know it has the form y = mx + b. And you have an m from solving the derivative. And you have an x and y, points it passes through. The only variable you don't have is b.
6 = 6 * 3 + b
And you solve for b.
6 = 18 + b
- 12 = b
And you put that back into your y = mx + b for the tangent line:
y = 6x - 12.

- anonymous

Thanks! woW, WHAT A RELIEF.

- anonymous

Since you seem to know about this stuff, do you know how to differentiate? This problem has been breaking my head for hours:
f(x)= \[f(x)= (x ^{2}-4x+2)(5x ^{3}-x ^{2}+5)\]

- anonymous

The whole thing f(x) is a product of two parts. So you should use the product rule.
f ' (x) = (the derivative of the 1st part) * (the 2nd part) + (the derivative of 2nd part) * (the first part).
The 1st part = x^2 - 4x + 2
The 2nd part = 5x^3 - x^2 + 5
You find the derivatives for the 1st and 2nd parts by using the power rule like for the other problem. Are you cool with how to do that or would you like help?

- anonymous

Yes i know I have to do that, but I keep missing up and I don't get the answer. Help me please :)

- anonymous

messing up*

- anonymous

OK. I'll the 1st and let you try the 2nd?
So for the first part, the first term is x ^ 2. The coefficient in front of the x^2 is 1. So when I bring the exponent down and multiply it by the coefficient, it is 2 x 1 = 2. The new exponent is the previous power, 2, minus 1. So 2 - 1 = 1.
For the next part, -4x, the derivative is just -4.
And for the constant at the end, +2, constants don't change so the derivative of that is 0.
To get the entire expression for the derivative of the 1st part, add up those parts.
Derivative 1st part = 2x - 4 + 0.
Hopefully that makes sense.

- anonymous

Is the second one this: \[15x ^{4}-2x+0\]

- anonymous

well without the x for the #2,

- anonymous

For the first term, you brought the 5*3 right. But then you added to the exponent, 3, instead of subtracting from it.
It should look like
15x^2 - 2x + 0.
Lemme play with the equation editor thingbob...
\[15x^2 - 2x + 0\]

- anonymous

Is that what I do now?
(2x+4)(15x^2+2x) + (15x^2+2x)(2x+4)

- anonymous

f ' (x) = (deriviatve 1st part)*(2nd part) + (derivative 2nd part)*(1st part)
1st part = x^2 - 4x + 2
2nd part = 5x^3 - x^2 + 5
Derivative 1st part = 2x - 4
Drivative 2nd part = 15x^2 - 2x
So:
f '(x) = \[f'(x) = (2x - 4)(5x^3 - x^2 + 5) + (15x^2 - 2x)( x^2 - 4x + 2)\]
Nasty to simplify but it is all algebra from here. :D

- anonymous

I think the algebra works out to:
Expand...
\[10x^4 - 2x^3 + 5x - 20x^3 + 4x^2 - 20 + 15x^4 - 60x^3 + 30 - 2x^3 + 8x^2 - 4x\]
Combine like terms of the same power:
\[ 25x^4 - 84x^3 + 12x^2 + x - 20\]
Hopefully all the algebra came out in the wash.

- anonymous

I GOT IT! lol sorry i was trying to solve it, but it took me forever! :)

- anonymous

Your really good at this!!!

- anonymous

Thanks. You'll be too, it just takes some patience and getting used to. Then calc is fun.

- anonymous

LOL I wish it was fun, I've been at it for days. I just dioscover this website today

- anonymous

Me too. It looks really fun but I don't know anyone here. :D
I have to go soon but it has been real nice talking with you!

- anonymous

They look like they really know their stuff here.

- anonymous

You do too! Good Night :)

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