rectangle inscribed in triangle

[rmn]

domain question> http://i207.photobucket.com/albums/b...EvErSe_/16.jpg
triangle question> http://i207.photobucket.com/albums/b..._/untitled.jpg
NOTE : PLEASE IGNORE THE NUMBER 1 MARKED ON TOP OF THE RECTANGLE IN THE TRIANGLE QUESTION (I DREW THAT).
i have the domain one but i'm not sure if I did it right, i got domain is (negative infinity, positive infinity) and my graph is going up with a top and then curves down like an upside down u and then for the secod question part a i got P(x,1/2) and part b i cant figure out the lengths of the rectangle ? if anyone could help it would be much appreciated thank you in advance. im just really worried ><

 

[stapel]

Your images aren't displaying. Please use the actual URL, not the "..." abbreviation. Thank you.

Eliz.

 

[rmn]

http://i207.photobucket.com/albums/bb23 ... Se_/16.jpg
http://i207.photobucket.com/albums/bb23 ... titled.jpg

 

[stapel]

I'm going to guess that you mean the following:

1) Find the domain of f(x) = 1 - 2x - x², and graph.

2) Consider a rectangle inscribed in an isosceles right triangle ABC, with A = (1, 0), C = (-1, 0), and B on the y-axis, with the right angle at B. Let the width of the half of the rectangle in the first quadrant be "x", so the rectangle's upper right corner meets the triangle on side AB at point P, P having x-coordinate "x".

a) Express the y-coordinate of P in terms of x. (Hint: Start by finding the line equation for AB.)
b) Express the area of the rectangle in terms of x.

1) The domain is any allowable x-values. Are there any forbidden x-values for this function? Are there any values of x that you can't plug in? The domain is everything else. As for the graph, where are you stuck?

2) You have the picture already, so you don't need to draw that.

a) Consider the right-hand half of the original triangle. You know the length of the base, and you know that the triangle is a right triangle. You also know, from the original triangle, what is the measure of the angle at A. From this, find the y-coordinate of B. Once you have this, find the slope of AB, and then the equation of the line through A and B.

Once you have the line equation, you have a formula for y, namely, "y = mx + b", for every point on that line segment (including P). Use that formula.

b) Area is given by height (y) times width (2x). Multiply.

If you get stuck, please reply showing your work and reasoning so far. Thank you!

Eliz.

 

[rmn]

the thing is is that we arent allowed to use calculators so how would i find tan 45?

for the first one i got (-infinity,+inifinity) and for the the second part i got a upside down parabola..am i doing it right? or shouldi try it again?

and is the height of the rectangle half the the triangle and the width half the triangle? it says the point is x but how can i find an area of x?

 

[Mrspi]

the thing is is that we arent allowed to use calculators so how would i find tan 45?

You should have learned some "right-triangle trig."

Draw a right triangle, with one acute angle having a measure of 45º. The other acute angle will be 45º also, right?

Now, if two angles of a triangle are equal, then the sides opposite these angles are equal, too. (This is often termed the Converse Isosceles Triangle Theorem.)

So, whatever the length of one leg (side opposite one 45º angle) is in this right triangle, the leg opposite the OTHER 45º angle has the same length.

tan 45º = opposite leg/adjacent leg

The two legs have the same length....so,

tan 45º = 1

YOU NEED TO MEMORIZE the functions for 45º, 30º and 60º. Yes....you can always "figure them out," as we did here....but that takes WAY too much time.

 

[stapel]

the thing is is that we arent allowed to use calculators so how would i find tan 45?

Where do you need the tangent? Why not just use the algebra, as explained above?

for the first one i got (-infinity,+inifinity)

I will guess that you mean, "for exercise (1), I believe the domain of the function is (-infinity, +infinity).

...and for the the second part i got a upside down parabola..am i doing it right? or shouldi try it again?

Since we cannot see your graph, and no information is provided about your intercepts, vertex, etc, we cannot know whether it is correct. But, yes, the general shape would be that of an "upside-down" parabola.

is the height of the rectangle half the the triangle and the width half the triangle?

No. The height is, as expained, the y-value of the point P. That's why you need to follow the provided steps and find the formula for y in terms of the half-width x. The height or width could be half those of the triangle, but this will vary with the value of x, which is not fixed.

it says the point is x but how can i find an area of x?

No; the exercise shows the half-width is given by x, the x-coordinate of the point P. Coordinates do not have area, and you are not asked to find "the area of x". You need to find the area of the rectangle.

Eliz.