Parabolas and dimensions?

Question

Hey guys i cant graph parabolas i keep messing up, and my finals tommarrow i also cant find coordinates of vertex and dont have any idea on how to do this problem A RECTANGLE HAS AN AREA OF 260 SQ INCHES. THE WIDTH IS 7 LESS THAN THE LENGHT. WHAT ARE THE DIMENSIONS OF THE RECTANGLE?
please tell me how to do these step by step

thanks a lot guys

Answer 1

The area equals width times length: A = W*L = 260
The width is 7 less than the length: W = L-7
A = W * L = (L-7) * L = L^2 - 7L = 260
L^2 -7L - 260 = 0
Solve using the quadratic equation (follow the link); the positive solution is L = 20. The width W = L-7 = 20-7 = 13.

Answer 2

I havent graphed parabolas by hand for a while, but this is how i think you graph them.
First, you find the vertex, it can be found at the point ( ( ( -b ) / ( 2a ) ), f( ( ( -b ) / ( 2a ) ) ) ), where f(x) is the function of the parabola.

Once you have found the vertex, plug in the x value of the vertex + 1 into the equation, and the x value of the vertex -1, then plot those three points on a graph, and connect them with a parabola graph.


The problem:
Area of a rectangle is A = Length * Width, the Length is x, and the Width is x - 7. We also know the area is 260 in^2.
A = Length * Width
260 = (x) (x - 7)
260 = x^2 - 7x
x^2 - 7x - 260 = 0

Here you can solve it whatever way you prefer...i will use the quadratic equation.
a = 1
b = -7
c = -260
x = ( - b +/- sqrt( b^2 - 4ac ) ) / (2a)
x = ( - (-7) +/- sqrt( (-7)^2 - 4(1)(-260) ) ) / ( 2(1) )
x = ( 7 +/- sqrt( 49 +1040 ) ) / 2
x = ( 7 +/- sqrt( 1089 ) ) / 2
x = ( 7 +/- 33 ) / 2
x = ( 7 + 33 ) / 2, x = ( 7 - 33 ) / 2
x = 40 / 2, x = -26 / 2
x = 20, x = -13

so, x = 20, because it cant have a negative length or width.

Length = 20 in.
Width = 20 - 7 = 13 in.

Answer 3

Call length =x
width =x-7
so
x(x-7)260
x^2-7x-260=0 2nd degree equation so
x=((7+-sqrt(49+1040))/2 = (7+-33)/2
As x>0 x=20inch = length
width=13inch