Intersection between two graphs
- Thread starter ManyTimes
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BigGlenntheHeavy
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\(\displaystyle Solve \ the \ system.\)
arthur ohlsten
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F(xy)=s,-s^2+4s
then:
x=s
y=-s^2+4s substitute x for s
y=-x^2+4x a parabola; completee the square
y=-[x^2-4x+4]+4
eq 1) y=-[x-2]^2 +4 a parabola open down, vertex at 2,4
===========================================================
S(x,y)= 8-4t, t
then:
x=8-4t
y=t substitute
x=8-4y
4y=-x+8
eq 2) y=-1/4 x +2 a straight line slope -1/4 y intercept=2
============================================================
you should be able to determine the two intercept points:
ste eq1) = eq 2) and solve for x
-[x-2]^2+4=-1/4x+2 solve for x , then substitute into either equation to find y
Arthur
then:
x=s
y=-s^2+4s substitute x for s
y=-x^2+4x a parabola; completee the square
y=-[x^2-4x+4]+4
eq 1) y=-[x-2]^2 +4 a parabola open down, vertex at 2,4
===========================================================
S(x,y)= 8-4t, t
then:
x=8-4t
y=t substitute
x=8-4y
4y=-x+8
eq 2) y=-1/4 x +2 a straight line slope -1/4 y intercept=2
============================================================
you should be able to determine the two intercept points:
ste eq1) = eq 2) and solve for x
-[x-2]^2+4=-1/4x+2 solve for x , then substitute into either equation to find y
Arthur
Thanks!!! Lovely!!!
But I still suck at this.
Did what you did and then solved the equation (hopefully it is correct).
What I've done:
-(x-2)² +4 = -1/4x+2
-x²-4x-4+2 = -1/4x
-x² + 6 = 3.75x //moving -4x over
6/3.75 = x²
sqrRoot(6/3.75) = x
1.265 = x
Then you ask me to do this:
" then substitute into either equation to find y"
And I did this:
y = -1/4*1.265+2
y = 1.684
And with the other equation I got this:
y = -(1.265-2)² + 4
y = 3.460
If I put it into the F(xy); x = 1.265, y = -1.265²+4*1.265 = 3.460
What does this tell me? Where do they intersect?
I know they should intersect two places. Around X= 6.80 and 1.2, then y is around -19 and 3.3, or maybe my calculator sucks? (I do know it sucks, it thinks that -2² = -4, which it is not)
No idea what im doing...
But I still suck at this.
Did what you did and then solved the equation (hopefully it is correct).
What I've done:
-(x-2)² +4 = -1/4x+2
-x²-4x-4+2 = -1/4x
-x² + 6 = 3.75x //moving -4x over
6/3.75 = x²
sqrRoot(6/3.75) = x
1.265 = x
Then you ask me to do this:
" then substitute into either equation to find y"
And I did this:
y = -1/4*1.265+2
y = 1.684
And with the other equation I got this:
y = -(1.265-2)² + 4
y = 3.460
If I put it into the F(xy); x = 1.265, y = -1.265²+4*1.265 = 3.460
What does this tell me? Where do they intersect?
I know they should intersect two places. Around X= 6.80 and 1.2, then y is around -19 and 3.3, or maybe my calculator sucks? (I do know it sucks, it thinks that -2² = -4, which it is not)
No idea what im doing...
mmm4444bot
Super Moderator
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- Oct 6, 2005
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- 10,962
ManyTimes said:
After you squared (x - 2), you did not properly distribute the factor of -1.
I suggest that you try completing one step at a time versus four steps at once.
(1) Square (x - 2)
(2) Distribute the factor of -1, to get rid of the parentheses
(3) Combine like terms
(4) Subtract 2 from both sides
Carry on.
-2^2 = -4 because we follow the Order of Operations (exponentiation before multiplication, yes?).
In other words, 2 gets squared first. Only then do we multiply by the factor of -1.
What you might be thinking of is: (-2)^2 = 4. (Note the parentheses. Grouping symbols are often used to change the order of some operations.)
In the case (-2)^2, both factors (-1)(2) get squared.