SOLUTION: I can not locate given function and locate intercepts of the function can someone help me with these problems? 1. Given the function f(x)=-6x^2-12x-2, what is the domain of f?

Question 174654: I can not locate given function and locate intercepts of the function can someone help me with these problems?
1. Given the function f(x)=-6x^2-12x-2, what is the domain of f?
A. real numbers b. {x\x>-1} the sign in the middle has a line under the lessthan. and a straigt line up and down at between the X's.
C. {x\x>or equal 1} or D. {x\xgreter or equal to-1} I choosen C.
2. Use a graph utility to graph the function over the indicated interval nad approx any local maxima nd local minima. If necessary, round answers to two decimals: f(x)=2+8x-x^2; (-5,5) here are the choices I have to pick:
a. local minimum at (4,50) B. local minimum at (4,18)
C. local minimum at (-4,18) D. local minimum at (-4,50)
I choosen A is that correct or where did I go wrong?
3. Find the average rate of change for the function between the given value:
f(x)= sqrt over 2x-1; from 1 to 5
A. 1/2 B. -28 C. -1/6 D. -2 I choosen B. is that correct?
4. Locate any intercept of the function.
f(x) = { on the side of -3x+9 and underneath 9x-3 if x<1 and underneath if v>1 with a line under the >
The answer if have to choose from is:
A. (0,3) B. (0,3),(3,0),(1/3,0) C. (0,9),(3,0),(1/3,0) D. (0,9)
I have choosen D is that correct..
I need some help and cannot find these in the book...

Answer by nycsub_teacher(90) (Show Source): You can put this solution on YOUR website!
Please list your questions one at a time instead of all questions in one post.
I will the following question below.
3. Find the average rate of change for the function between the given value:
f(x)= sqrt over 2x-1; from 1 to 5
A. 1/2 B. -28 C. -1/6 D. -2 I choosen B. is that correct?
=========================================================
Hello. I will go over the question again SLOWLY. Perhaps I made a mistake. If so, I'll let you know along the way. You said we need to "square over." I assume (2x - 1) lies in the radicand, right?

I will assume that you are talking about the square root of the quantity (2x - 1), which is written as sqrt{2x - 1}.

The average rate of change formula = [f(b) - f(a)]/(b - a)

You were given a function over the interval [1, 5], which means from 1 to 5.

In this case, a = 1 and b = 5.

We now search for f(b), then we search for f(a).
Once we have those two guys, we plug them into the formula above and simplify.

Are you with me so far?

Remember, b = 5 and a = 1 from the given interval.
We plug each number into the function and simplify. By doing so, we will find f(b) and f(a).

Here it is:

f(x) = sqrt{2x - 1}

f(5) = sqrt{2(5) - 1}

f(5) = sqrt{10 - 1}

f(5) = sqrt{9}

f(5) = 3

This means that f(5) = 3 = f(b)

I see what happened. I made a mistake.

The square root of 9 is 3 not 9.

So, f(b) = 3 NOT 9.

Sorry about that....

==========================

We now find f(a).

f(1) = sqrt{2(1) - 1}

f(1) = sqrt{2 - 1}

f(1) = sqrt{1}

f(1) = 1 = f(a).

==========================

We can now plug that info into the formula and simplify.

average rate of change = (3 - 1)/(5 - 1)

average rate of change = 2/4 = 1/2

Dawn, I apologize. The correct answer is 1/2.

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