An equation of a line parallel to y=x-6, must have the same slope.
In this equation:
y=mx+b (slope-intercept form)
m is the slope:
The slope of the equation y=x-6 is m=1 (the number beside "x").
Now we have a point (-1,5) and the slope m=1.
Point-slope form of a line:
y-y₀=m(x-x₀)
so:
y-5=1(x+1)
answer: the equation of the line in point-slope form is :
y-5=1(x+1)
And the eqution of this line in slope-intercept form is:
y=x+6
y-5=(x+1)
y=x+1+5
y=x+6
Step-by-step explanation:
TO FIND THE DOMAIN OF THIS FUCTION AVOID SITUATIONS LIKE Y=2/0 FOR THIS IS UNDEFINED.
TAKING THE DENOMINATOR AND EQUATING IT TO ZERO
=> X – 6 =0
=> X=6
SO AVOID HAVING X = 6 IN THE DENOMINATOR FOR WHICH THE FUNCTION WILL BECOME UNDEFINED
THEREFORE, X ≠6
DOMAIN ={ X£R, x ≠6}
I.e all real numbers except 6
The x-axis.
In the form Y = mx + b
m is the slope, so no slope means that my is zero so that means x is zero also.
So you would end up with
y = b so that means y will always be that value and the only thing that will change is the x value. This will create a horizontal lines that is parallel to the x-axis
f(g(-1)) = - 3
Evaluate g(-1) and substitute into f(x)
g(-1) = (-1)² -7(-1) - 9 = 1 + 7 - 9 = - 1
f(g(-1)) = (-1) - 2 = - 3
