Quadratic Inequalities?

Question

Solve the quadratic inequality for x:
-6x^2 + 18x + 24 ≤ 0
What is the MAXIMUM number of zeros for a quadratic function:
y = ax^2 + bx + c
What is the MAXIMUM value for the following quadratic function?
A rectangular pen is to be constructed using 280 feet of fencing material. What is the maximum area possible for such a pen if the pen is to be constructed adjacent to a barn? (hint: there are only 3 sides of fence)
What are the dimensions for the pen in the previous problem?
A cumquat is propelled directly upward from ground level with an initial velocity of 50 feet per second. Let t represent the amount of time elapsed since the cumquat is thrown upward. Write the function S(t) for the height of the cumquat as a function of time t
15.A ball is tossed directly upward with an initial velocity of 60 feet per second from 6 feet above the ground. The height of the ball, s, as a function of time, t, is given by the formula, s = -16t^2 + 60t + 6.
When is the ball back onon the ground? (How many seconds from toss to ground

Answer 1

-6x^2 + 18x + 24 <= 0

One thing to note is that all of the coefficients are divisible by 6. Let's divide the whole equation by -6 (switching the inequality sign doing so, since multiplying/dividing by a negative number does that)

x^2 - 3x - 4 >= 0

Now, we factor normally

(x - 4) (x + 1) >= 0

Our critical values are -1 and 4. We want to determine when the inequality is true, by testing three intervals: numbers less than -1, numbers between -1 and 4, and numbers greater than 4.

First, let's test a number less than -1: -1000000. Then we get:
(negative) times (negative) = positive, so we include the interval (-infinity, -1)

Next, let's test a number between -1 and 4; 0 is a good one. So we get (0 - 4) (0 + 1) = negative times positive = negative. So we don't use this interval.

You'll find that a number greater than 4 will yield a positive result too.

Therefore, our answer for x is
x is an element of (-infinity, -1] U [4, infinity)

Note that I put the square brackets, because -1 and 4 *are* included. They make the function equal to zero, and our sign indicates "greater than OR equal to".

2) The maximum number of zeros for a quadratic function is 2. If you take higher levels of math, this is known as the Fundamental Theorem of Algebra.

3) Since we have 280 feet of fencing material, and the pen is to be constructed adjacent to a farm, that means we have three sides (with the 4th side being the barn). The pen is rectangular, so let's call P the perimeter of the there sides. Then, if we let L be the length and W be the width,

P = L + L + W, or
P = 2L + W

But we're given that P = 280 (there's 280 feet of fencing material), so

280 = 2L + W

What we want to maximize is the area, and we actually have a formula for the area of the pen:

A = LW

But, we can express W as a function of L, since
280 = 2L + W
280 - 2L = W

Therefore,

A = L(280 - 2L)
A = 280L - 2L^2

Therefore, since we now have one variable, we can call this our function, A(L).

A(L) = 280L - 2L^2

To find the maximum area, we find A'(L) and then make it 0.

A'(L) = 280 - 4L
Therefore
0 = 280 - 4L
4L = 280
L = 70

The maximum possible area is found by calculating A(70).

A(70) = 280(70) - 4(70)^2 = 9800 feet^2.

As for the dimensions, we just solved for the length, L, which is 70. The width we can calculate from W = 280 - 2L = 280 - 2(7) = 140.

L = 70, W = 140

Answer 2

Do you know how to solve a quadratic equality? If so you basically will do that. Then determine the 2 #s that add to -1 (the coefficient next to the x) and multiply to 6. (-x + a)(x + b): find a and b. (they are -3 and 2 respectively) -x^2 - x + 6) = (-x - 3)(x - 2) Now you have your values for x where they cross the x axis. aka -x-3 = 0 therefore x = -3. When x = -3 the equation = 0 which means it intersects the x-axis at x = -3. Same goes for x = 2. So far everything has been solved just as it would be for an quadratic equality, right?! Right!! Now you have to determine what x values would cause the equation to be <0, you know x cannot equal 2 or -3 since those equal 0 not less than 0; however you can use those two values as a starting point. Use x = -4 in the equation, does that give a value less than 0? -(-4^2) -(-4) + 6 = -6, so yes it does equal less than 0. Now you know that x can be any number < -3. Use x = 0 (value between -3 and 2) -(0^2) - 0 +6 = 6, this is not <0 therefore you now know the x value cannot be between -3 and 2. Use x = 3 -(3^2) - (3) + 6 = -6, so yes once again this equals less than 0. Now you know x can be anything > 2. So your final answer would be x > 2; x < -3. Hope that helps!!

Answer 3

1) -6x^2+18x+24 ≤ 0
-6(x-4)(x+1) ≤ 0
(x-4)(x+1) ≤ 0
Therefore: x>=4 or x<=-1. In other words, any number less than -1 or greater than 4 will satisfy this inequality.

2)The maximum number of zeros in a quadratic is always 2. Think about it this way, whats the maximum number of times a parabola can cross the x-axis.

3+4) See previous poster, who answer it correctly.

5)S(t)=50*t

6)s(t)=-16t^2 + 60t + 6.
We want to know when s(t)=0.
So we will need to solve:
-16t^2 + 60t + 6 = 0
You need to use the quadratic formula here since this polynomial cannot be factored into two linear parts. The answer I got was:

t=3.85 and t = -0.097

Notice that the second answer doesn't make any sense since we are not concerned with what the comquat does before t=0. Therefore the answer is: the comquat hits the ground at t = 3.85 sec.

Cheers!

Answer 4

Solve the quadratic inequality for x:
-6x^2 + 18x + 24 ≤ 0

(-3x+12)(2x+2) ≤ 0

= (x:I -3 ≤ x ≤ -1 )

That's the first question. Good luck on the secound part.