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Question 1077703: Hi, I'm working on an algebra project and I have a couple of questions that I need help with. The first question is: Write an expression for the width of any rectangle in terms of the length. The second question is: Write a function for the area of any rectangle in terms of the length. Also on this last question I need help on how to graph it and to find the vertex. Thank you!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x equal the length of the rectangle
let w equal the width of the rectangle.
let y = the area of the rectangle.
area of a rectangle = length * width.
this translates to:
y = x * w
let k = the length of the rectangle minus the width of the rectangle.
you get k = x - w
solve this equation for w to get w = x - k
your equation for the area of the rectangle was:
y = x * w
replace w with x - k and the equation becomes:
y = x * (x - k)
simplify this equation to get:
y = x^2 - kx
this is a quadratic equation.
if we set y = 0, then it becomes:
x^2 - kx = 0
now it is in standard form of ax^2 + bx + c = 0, where a = 1 and b = -k and c = 0
the x-coordinate of the vertex of this equation is at x = -b/2a.
this becomes x = -(-k)/2) which becomes x = k/2.
the y-coordinate of the vertex of this equation is at y = f(-b/2a) which becomes y = f(k/2).
you solve for f(-b/2a) = f(k/2) by replacing x in the equation of y = x^2 - kx with (k/2)
the equation becomes y = (k/2)^2 - k*(k/2)
simplify this to get y = k^2/4 - k^2/2
place all fractions under a common denominator to get y = k^2/4 - 2k^2/4.
combine like terms to get y = -k^2/4.
your vertex is therefore at (k/2, -k^2/4)
what does this mean in real terms?
let's see:
assume the length of your rectangle is 7 and the width of your rectangle is 3.
since we set the equation up as y = x * w, where y is the area, x is the length, and w is the width, then we get:
y = x * w becomes y = 7 * 3 which becomes y = 21 which is the area of the rectangle.
all well and good, except that we want the equation in terms of x.
we did that earlier and we got:
y = x^2 - kx, where y is the area, x is the length and k is equal to x - w which is the length minus the width.
when x = 7 and w = 3, we get k = 7 - 3 = 4
the equation of y = x^2 - kx becomes y = x^2 - 4x
if we set y = 0, the equation becomes:
x^2 - 4x = 0
since this is a quadratic equation in standard form, we get:
a = 1
b = -4
c = 0
the x-coordinate of the vertex is at x = -b/2a which becomes x = 4/2 which becomes x = 2
the y-coordinate of the vertex is at f(-b/2a) which is at y = 2^2 - 4*2 which is at y = 4 - 8 = -4.
the vertex is therefore at (2,-4)
you can graph the equation of y = x^2 - 4x and it will look like this:
in this graph, y is the area and x is the length.
the width is assumed to be equal to the length minus 4.
what this means is:
when the length is 4, the width is 0
when the length is 5, the width is 1
when the length is 6, the width is 2
when the length is 7, the width is 3
your area is equal to y.
you will get:
y is equal to 0 when x is equal to 4
y is equal to 5 when x is equal to 5
y is equal to 12 when x is equal to 6
y is equal to 21 when x is equal to 7
vertical lines are placed on the graph to show you the value of y when x is the value indicated by the vertical line.
that graph is shown below:
what this says is that, as long as you know the value of k, you can find the area of the rectangle in terms of the length.
since we know the value of k, we can test this out.
the following graph shows the vertex as well as the value of y when x = 7.
we know that the value of k is equal to 7 - 3 which is equal to 4.
so we have k = 4.
and we have y = x^2 - kx which becomes y = x^2 - 4x when k = 4.
we also determined that the x-coordinate of the vertex of the equation would be at k/2, and we also determined that the y-coordinate of the vertex of the equation would be at -k^2/4
the k = 4, we get:
x-coordinate of the vertex of the equation is at 4/2 = 2
y-coordinate of the vertex of the equation is at -4^2/4 = -16/4 = -4.
since we already showed that the vertex of the equation is at (2,-4), then these values based on k = 4 are correct.
to summarize.
y is the area.
x is the length
k is the value of the length minus the width.
the equation becomes y = x^2 - kx.
the x-coordinate of the vertex of the equation is x = k/2
the y-coordinate of the vertex of the equation is y = -k^2/4
this should work with any value of length and width.
let's assume, arbitrarily, that the length is 50 and the width is 75.
this means that k = 50 - 75 = -25.
y = x^2 - kx becomes y = x^2 + 25x
x-coordinate of the vertex becomes y = k/2 = -25/2 = -12.5
y-coordinate of the vertex becomes y = -k^2/4 = -156.25
when the length is 50, the width is 75 and the area is equal to 50 * 75 = 3750.
all of this can be seen on the following graph of the equation of y = x^2 + 25x.
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