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do you know what \(y=(x+1)(6+3x)\) looks like?

a parabola

upward

a parabola that opens up, right
and the zeros are?

-1 and -2

ok good so lets draw a quick picture

|dw:1441157312300:dw|

no it has a relative math right at \(x=-2\)
by a similar argument it has a relative min at \(x=-1\)

you lost me..

ok lets go slow then

you know what the derivative looks like right?

yes

yes i learned that

ok good
now "positive" is a synonym for "above the x axis"

|dw:1441157806983:dw|

that means on the interval
\[(-\infty,-2)\] and \[(-1,\infty)\]the function is increasing

not the derivative, the original function \(f\)
also on \((-2,-1)\) the function is decreasing

Ohh okay so you just us eincreasing/decreasing rules/ i understand thanks

yw

Can you help me with my second question as well?

sure but i didn't see it

oh now i do
take the derivative
what do you get?

i hope you get \[t^2-15t+50\]

For the derivative I got :t^2-50t+50

oops i meant 15

ok so your job is to solve \[t^2-15t+50<0\]

because if the derivative is negative, it is moving to the left

fortunately for you this factors, so it should be easy

t<5

is it clear how to solve that, or no?

t>10

i think you solved for where it is positive

How am I supposed to solve this then? The two intervals i got were those

it is zero at 5 and 10 right?

|dw:1441158937764:dw|

yes i got this

is moving left another name for negative?

moving to the left = derivative is negative

i.e. the velocity is negative