:/

y=x^2+6x+1 find the domain and range.

all read numbers and i will give you a hint for the range i would write f in vertex form and also we know the function f is concave up so we have the range is [the y part of the vertex, infinity]

x is all real numbers f(x)>-8

x is all read numbers

lol

that d was suppose to be an l

lol

thats so funny

its not that funny :)

\[f(x) = x^2+6x+1\]\[f'(x) = 2x + 6\] \[f'(x) = 0\]\[2x + 6 = 0\]\[ x = -3\] \[f(-3) = (-3)^2 + 6(-3) + 1 = -8\] Range: \([-8, \infty)\)

calculus way is funner

And easier!

yep

i don't understand how you guys got that answer

the one with the weird symbol?

yup

this symbol ' ?

don't worry about that way

We got it with Calculus!

ohhhhh. good thing i didn't write that down

lol

will you take cal?

when I'm a junior in high school I will.

you will find that calculus is more fun than algebra

You CAN write it down. Impress your teachers!!

;)

For algebra you just have to see that it's a parabola opening upwards, so the minimum value will be the vertex. So find the vertex, and the range will be \([y_1, \infty)\) where \(y_1\) is the y value of the vertex.

hopefully it will be more fun. and hahaha, my teacher would probably freak out.

no what would be more impressive if you actually used the formal definition of derivative

but how do you know that it's a parabola pointing upwards in the first place?

Well I think writing it down without understanding it will only impress upon h(is/er) teacher that (s)he cheated.

f=x^2 is concave up f=-x^2 is concave down

Because the leading coefficient is positive.

ok. and i also have to graph this, and i get how to tell if its upwards/downwards now, but how do i know how wide the arc is?

f=cx^2 c>0=> concave up c<0=> concave down

that probably didn't make much sense..

Well you can solve it for some y value (say y=0) and then you'll get two x values. That'll give you a decent approximation for a sketch.

ok. and another question is \[y=3(2)^{x}\] can I just plug in 0 as the x exponent?

What you are really looking at is the limiting behaviour of the function. It is because the leading term dominates the other terms as \(x\) approaches infinity that you are worried about the sign of its coefficient.

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