1. ## Quadratics Need Help Asap :/

If anybody could help me with the equations and/or solutions to one or more of these problems, it would be greatly appreciated.. I'm completely lost

- As an employee of a landscaping company, you have been asked to create a rectangular flowerbed, which fits these specifications. You must use two twelve foot pieces of timber to outline the bed. The bed must have an area of 32 square feet. Create a quadratic equation to represent the situation. Solve the equation with one algebraic method and use the solution(s) from the equation to describe where to cut the timbers. Show all work and discuss all processes.

-The next customer that your landscaping company is working for has a similar situation. They owuld like to create a bed out of two twelve foot pieces of timber but would like to maximize the area that is created from these timbers. Create a quadratic equation to represent the situation. Solve the equation with one algebraic method and use the solution(s) from the equation that was created to describe where to cut the timbers. Show all work and discuss all processes.

-Write two quadratic functions whose graph has a vertex in the second quadrant. One must have no real roots and the second must have two real roots. For each function find the vertex, line of symmetry, and roots (real or complex). Justify why one of the function has no real solution and the other has two real solutions. Show all work and discuss all processes.

THANK YOU
-Rachel

2. Originally Posted by armthezombies
As an employee of a landscaping company, you have been asked to create a rectangular flowerbed, which fits these specifications. You must use two twelve foot pieces of timber to outline the bed. The bed must have an area of 32 square feet. Create a quadratic equation to represent the situation. Solve the equation with one algebraic method and use the solution(s) from the equation to describe where to cut the timbers. Show all work and discuss all processes.
You are building a rectangle, so call the length x and the width y. You have 2 12 foot long boards, so the perimeter of the rectangle must be 2*12 = 24 feet. The area of the flowerbed is 32 square feet. Thus:
2x + 2y = 24
x*y = 32

So solve the top equation for, say y, and then plug that into the bottom equation:
$y=12-x$
$xy=x(12-x)=32$

So solve the quadratic for x, plug that into your y equation and that will tell you how to cut each of the 12 foot boards. (I got x=8 feet and y=4 feet, or x=4 feet and y=8 feet. They both represent the same thing.)

-Dan

3. Originally Posted by armthezombies
The next customer that your landscaping company is working for has a similar situation. They owuld like to create a bed out of two twelve foot pieces of timber but would like to maximize the area that is created from these timbers. Create a quadratic equation to represent the situation. Solve the equation with one algebraic method and use the solution(s) from the equation that was created to describe where to cut the timbers. Show all work and discuss all processes.
Same basic process as the last problem. So I'll skip to almost the end.

Given x and y as dimensions of the flowerbed, the area is:
$A = 12x-x^2$
We need to find a value of x such that A is a maximum value. The point here is that the expression $12x-x^2$ is parabolic. So what do we know about the graph of $y=-x^2-12x$? It is a parabola opening downward, so the vertex point will be the maximum y value for the graph. This is the point you are looking for.

-Dan

4. Originally Posted by armthezombies
Write two quadratic functions whose graph has a vertex in the second quadrant. One must have no real roots and the second must have two real roots. For each function find the vertex, line of symmetry, and roots (real or complex). Justify why one of the function has no real solution and the other has two real solutions. Show all work and discuss all processes.
Given a parabola $y=ax^2+bx+c$ we know that the axis of symmetry will be $x=-b/(2a)$ and the vertex is either the highest or lowest point on the graph and will be at that value of x. You can get the roots by using the quadratic formula. So all we need to do is decide on a, b, and c for the two cases.

Since you are looking for a vertex in the second quadrant, you need to pick an a and b such that $x=-b/(2a)$ is a negative number. Once you do that pick a c such that $y=ax^2+bx+c$ is a positive number.

As for the two real and complex roots, I'll let you figure that out. I'll only ask you the following question: What is it about the form $ax^2+bx+c$ that makes a parabola open upward or downward?

-Dan