Graphing

[Mrtinkles]

Hello, I am wondering if someone can help with part b on this. I'm not sure where to start with it

The position of an animal on a two dimensional map can be described by its x, y coordinates. You are given that the path of a lion can be described by the equation y = a × x + b. Similarly the path of a deer can be described by the equation y = c × x2 + d × x + e. Note that here a, b, c, d and e are constants.


(a) If a = 1, b = 5, c = 1/10, d = 1 and e = 1, plot these two equations on an x-y graph showing clearly the points of their intersection.

y=1x+5
y= 1/10x^2+1x+1

intersect (-6.3,-1.3) (6.3,11.3)


(b) For general a, b, c, d and e, find equations for the positions in space (the values of x and y) at which the deer and the lion meet.

 

[Mrtinkles]

How does this look?

m= y2-y1 / x2-x2
m =-1.3 - 11.3 / -6.3 - 6.3
m = -12.6 / -12.6
m = 1

y=mx+b
y=1x+b

-1.3 = 1.3 * -6.3 +b
b = -1.3 - (1.3)(-6.3)
b = 6.89

y=1x+6.89

 

[Harry_the_cat]

How does this look?

m= y2-y1 / x2-x2
m =-1.3 - 11.3 / -6.3 - 6.3
m = -12.6 / -12.6
m = 1

y=mx+b
y=1x+b

-1.3 = 1.3 * -6.3 +b
b = -1.3 - (1.3)(-6.3)
b = 6.89

y=1x+6.89

Why do you have -1.3=1.3*-6.3+b ? Why the 1.3? Shouldn't it be 1?
But then you will just get b=5. You already know that the equation of the line is y=1x+5.

 

[Harry_the_cat]

The question is asking you to go back to the general case (ie, a,b, etc are no longer 1, 5 etc) and find a formula or equation for the intersection points in terms of a,b,c,d,e. Your answer will be in the form x=.... and y= ....

 

[Mrtinkles]

Oh yes, it is meant to be 1 not 1.3.
Do you know a method for getting this or videos I can watch?

 

[Harry_the_cat]

Would you know how to algebraically find the points of intersection of the line and parabola in part (a)?

 

[Mrtinkles]

I think so, something like y=y, 1x+5 = 1/10x^2+1x+1 ?
I'm still a bit confused and not sure what format the answer wants to be in also, like x=(5,1) or like x=5 ?

 

[Mrtinkles]

I've found a video which seems to cover it but can you tell me if I'm using the right method?

This doesn't match what the computer says the intersect is when I work it out fully so maybe you can see what is wrong

y=y
1/10 x² + x +1 = x+5.......-5
1/10 x² +x -4 = x......-x
1/10 x² -4 = 0

(x-4)(x+1/10) = 0
x = +4, -1/10

 

[HallsofIvy]

You still are not doing what the problem told you to do! "(b) For general a, b, c, d and e, find equations for the positions in space (the values of x and y) at which the deer and the lion meet."

In other words, you want to find the values of x and y that satisfy y = ax+ b and y = cx^2 + dx + e.

So y= ax+ b= cx^2+ dx+ e. That is equivalent to cx^2+ (d- a)x+ (e- b). Solve that using the quadratic equation.

 

[Harry_the_cat]

You still are not doing what the problem told you to do! "(b) For general a, b, c, d and e, find equations for the positions in space (the values of x and y) at which the deer and the lion meet."

In other words, you want to find the values of x and y that satisfy y = ax+ b and y = cx^2 + dx + e.

So y= ax+ b= cx^2+ dx+ e. That is equivalent to cx^2+ (d- a)x+ (e- b). Solve that using the quadratic equation.

Should read:
That is equivalent to cx^2+ (d- a)x+ (e- b)=0

 

[Harry_the_cat]

I've found a video which seems to cover it but can you tell me if I'm using the right method?

This doesn't match what the computer says the intersect is when I work it out fully so maybe you can see what is wrong

y=y
1/10 x² + x +1 = x+5.......-5
1/10 x² +x -4 = x......-x
1/10 x² -4 = 0

(x-4)(x+1/10) = 0
x = +4, -1/10

You're correct up to \(\displaystyle \frac{1}{10}x^2-4=0\), but then you factorise it incorrectly.
\(\displaystyle x^2-40=0\)
\(\displaystyle (x-\sqrt{40})(x+\sqrt{40})=0\), etc.

BUT, the question is asking you to find x and y in terms of a,b,c,d and e. As Halls said, you need to solve
\(\displaystyle cx^2+ (d- a)x+ (e- b)=0 \).
Use the quadratic formula to solve for x.