The slope of line cannot be that of curve or greater.
slope line=-3
slope curve>slope line
-3<-3/x^2
x^2>1
x>1
check x=10 slope line=-3
slope curve=-3/100=-.03
check x=1/2 slope curve=-12
do the others the same way
1. Given that the straight line y=c-3x does not intersect the curve xy=3, find the range of values for c.
2. Find the range of values for c for which the line y=cx+6 does not meet the curve 2x^2-xy=3.
3. Find the range of values for k for which 8y=x+2k is a tangent to the curve 2y^2=x+k
4 answers
number one is supposed to be -6<c<6
number two is supposed to be c>5
and number three is supposed to be k=8
number two is supposed to be c>5
and number three is supposed to be k=8
the slope of xy=3 is y' = -3/x^2
So, the slope is -3 at x=1
The tangent to the curve at x=1 is
y-3 = -3(x-1)
y = -3x+6
y = 6-3x
so, for y = c-3x, if c < 6, the line falls below the curve. Similarly, for x<0, the line lies above the other branch of the curve if c > -6.
So, -6 < c < 6
Follow this logic for the other parts
So, the slope is -3 at x=1
The tangent to the curve at x=1 is
y-3 = -3(x-1)
y = -3x+6
y = 6-3x
so, for y = c-3x, if c < 6, the line falls below the curve. Similarly, for x<0, the line lies above the other branch of the curve if c > -6.
So, -6 < c < 6
Follow this logic for the other parts
Bruh i don't understand yallz explanation