Question

What is the equation of a line that passes through the point (4, 2) and is perpendicular to the line whose equation is y=x3−1 ?

Answer 1

I'm assuming that the equation y=x3 - 1 is a cubic = x^3 - 1
it's slope at x=4 is y'= 3x^2 so y'(4) = 48
Then the slope of the line perpendicular at the point (4,2) is = -1/48
So y - 2 = -(1/48)(x-4) = -x/48 + 1/12

Answer 2

So since we know the line is perpendicular to the line y = 3x − 1, it will have the same slope, 3. So we know our equation will be y = 3x + ?, that ? is the y intercept. To find this you can plot the point (4, 2) and graph a line with a slope of 3 that passes through that point. Then you see where that graphed line passes through the y axis, aka the y intercept. Or you can take the point (4,2) and for every 1 you subtract from the x position, subtract 3 from the y. This is because of our slope of 3. So... you start with (4,2) then (3,-1) then (2,-4) then (1,-7) then (0,-10). Finally you have the point (0,10) which is a point that is on our line, and is the y intercept. Therefore the equation of a line that passes through the point (4, 2) and is perpendicular to the line whose equation is y = 3x − 1 is y = 3x - 10