Question

points of intersection, for y=x^2+3x-4 and y=5x+11. would you eventually solve it as a quadratic?

Answer 1

simply set these two equations equal to each other like so:
x^2+3x-4=5x+11 and evluate

Answer 2

What Lagrange said.

Answer 3

yes...it will be a quadratic

Answer 4

yeah i started that, and ended up x^2= 2x+15 would i move it over to one side and solve it as a quadratic?

Answer 5

yep

Answer 6

what would i do for y=cosx and y=sinx x in the first quadrant? i have an idea but im not so sure

Answer 7

whats the question?

Answer 8

these are your x and y for the quadratic?

Answer 9

no no, that would be the next question the next is to determine points of intersection for y=cos x and y=sinx in the first quadrant

Answer 10

oh ic.

Answer 11

set them equal to one another

Answer 12

cosx=sinx
when is this true for \[0<x < \frac {\pi}{2}\]

Answer 13

oh my i honestly dont remember, hold on im looking throught notes, uhh

Answer 14

We can also look at this from the definitions of sine and cosine:
\[\frac{x}{r}=\frac{y}{r}\] so x=y. We thus have a triangle with hypotenuse=1 and sides of equal length...

Answer 15

this is known as an isosceles triangle. if you remember they have 2 angles of 45 degrees (one 90 degrees)

Answer 16

so sinx=cosx when x=45 degrees or pi/4 radians

Answer 17

so how would i show work for this, i see what ur trying to say, but at the same time im a bit lost

Answer 18

I would draw a right angle triangle with sides a=1, b=1 and c=sqrt(2). We know a=b from the definitions of the sin and cos functions as I said above. We want the unknown angle between the c and a (or b). The only way this can be true is if the angle is 45 degrees. Think of starting with a square of sides a=b. Now cut it into two equal right triangles. The only way to do this is to split the square along the diagonal. This divides angle ab (which is 90 degrees) in half, giving you a 45 degree angle.

Answer 19

sorry if this is confusing

Answer 20

its asking to solve the points of intersectionusing algebra :/ ive done that u are explaining so i understand that

Answer 21

what*

Answer 22

Then it is just sinx=cosx. You're probably expected to remember the sin and cos of common angles. So, from your table of common angles you cite x=pi/4 as the angle in the first quadrant at which sinx=cosx

Answer 23

oh ok.

Answer 24

so in that case it would be radical 2 over 2

Answer 25

Yeah. there is only one point where sinx intersects cosx in the first quadrant and that occurs at \[(\frac{\pi}{4},\frac{\sqrt{2}}{2})\]

Answer 26

i have no idea how to show that using algebra like the first problem, but i do know the answer which might get me points taken off lol

Answer 27

You can show it with geometry but not for algebra in the sense we did for the last problem

Answer 28

wait!

Answer 29

yeah i know thats why im wondering why it is asking me to solve it usng algebra.

Answer 30

I just thought of it.
One sec gotta put my kid to bed.

Answer 31

its ok no problem.

Answer 32

would you use trig identities.? im not sure i just happened to find them in my notes

Answer 33

okay. let r^2=a^2+b^2
we know (from the definition of the sine and cosine functions) that cosx=a/r
and sinx=b/r
We want sinx=cosx
so a/r=b/r
or, a=b
So we have r^2=a^2+a^2
r^2=2a^2
On the unit circle we have r=1. So,
1=2a^2
or, a=sqrt(1/2)
If we rationalize the denominator this becomes \[\sqrt{2}/2\]So, on unit circle sinx=cosx=(sqrt2)/2
Now, from your table of common angles we read that cosx and sine x both equal (sqrt2)/2 at x=pi/4

Answer 34

thank you so much so much and thank you so much for being so patient with me!

Answer 35

no sorry for the long response! usually im more on the ball. I got a 3 year old and a 14 month old in the background which doesn't help me focus lol

Answer 36

its alright, i didnt mind the long response gave me time to think about this problem more lol