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| 11 May 2015 06:38 PM |
Is there a tangent line for:
f(x) = x^3 + 6x^2 + 21x + 26?
When I did it, I got 3x^2 + 12x + 21 which has imaginary roots.
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Sandykinz
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ilezybet
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| 11 May 2015 06:43 PM |
8 x 3 = 24 + 1 =25 + 3 + 2= 30
Renamed2163391 - Acclaimed RTer |
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| 11 May 2015 06:43 PM |
| Well apparently I was correct for the derivative. That was an easy problem. |
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LegoADS
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| 11 May 2015 06:43 PM |
f(x) = x^3 + 6x^2 + 21x + 26?
is this an equation of a circle? or this is just a tangent with the radius of a circle??
a^2+b^2=c^2
so
7^2+21^2=26^2 i believe so if these are equal then it is a tangent
441 = 7^2+21^2 26^2 = 676
so i think the answer is no...
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LegoADS
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| 11 May 2015 06:45 PM |
| ok i have no idea what ive done i will have to look into this more |
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| 11 May 2015 06:46 PM |
| Yeah I plugged the equation into the quadratic formula and it had imaginary roots. |
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| 11 May 2015 06:46 PM |
| Also yeah you don't know what you are talking about lol. |
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LegoADS
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| 11 May 2015 06:49 PM |
ok so
'Both the graph of y = f(x) and the tangent line pass through the point, and the tangent line has the same gradient, 'm', as the function at that point.'
f(x) = x^3 + 6x^2 + 21x + 26
ok so y = m(x-x1)+y1
ok this is confusing so
y =
so basically you would have to work out the gradient and probably do a table of results for each x and y value and put that in the equation to find the point of the tangent |
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| 11 May 2015 06:54 PM |
yes it has "a tangent line" in fact it has infinite tangent lines, one at each value of x fail
~This sig is false~ |
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| 11 May 2015 06:56 PM |
@MathFreak, would it be:
x = +- i
y = +-21i +-i + 26 +-6
and therefore have (-infinity, +-i), (=-i, infinity) for the areas of increase and decrease of the equation? |
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LegoADS
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| 11 May 2015 06:57 PM |
so you pretty much have to work out the point the tangent meets the curve and then find the normal which is the perpendicular line to the tangent
like math said there should in theory be an infinite number of tangents that can meet the curve??? is there any extra info |
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LegoADS
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| 11 May 2015 06:59 PM |
o im 15 so ive covered like none of that
that looks beyond or roughly A-level standard |
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Sandykinz
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| 11 May 2015 06:59 PM |
| what is all this fuss about |
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| 11 May 2015 07:00 PM |
| Oh wait, isn't the slope for the max and min of the graph 0? |
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| 11 May 2015 07:13 PM |
if your derivative has all imaginary roots that simply means that the slope of the polynomial is never zero
but it still has a derivative
i'm not huge on calc yet but that's what i'm going for
~This sig is false~ |
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| 11 May 2015 07:26 PM |
| Yeah I got the derivative, but I have to find the max and min by using it. As well as the value and location. So all of my answers look completely weird. |
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| 11 May 2015 07:27 PM |
well like i said the derivative is never 0 so there are no mins/maxs
~This sig is false~ |
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