2006-09-21 01:15:49 · 5 answers · asked by mathsnoob 1 in Science & Mathematics ➔ Mathematics
If
y = 4x^3 - x^2 + 3
( you should find derivative of y )
y' = 12x^2 - 2x
x = 0
y' = 12(0)^2 - 2(0)
y' = 0
Good Luck
2006-09-21 01:37:11 · answer #1 · answered by sweetie 5 · 1⤊ 0⤋
The point where it crosses the y-axis is a point where x = 0. Thus, we need to find its derivative at that point.
y = 4x³ - x² + 3
The derivative is
dy = 12x² dx - 2x dx
Thus, dy/dx is
dy/dx = 12x² - 2x
When x = 0,
dy/dx = 12(0)² - 2(0)
Thus,
dy/dx = 0
Therefore, the slope of the tangent line to that curve at the y-intercept is 0. Therefore, the line is horizontal.
Since the normal line is perpendicular to the tangent line, the normal line is vertical. Thus, it is in the form x = h, where h is a value (constant for x).
Since it passes through a point in x = 0, then the equation for the normal line is x = 0.
^_^
2006-09-21 01:22:49 · answer #2 · answered by kevin! 5 · 0⤊ 0⤋
I think you should do your own homework. That is the purpose of homework, to teach you how to do it and make sure the lesson sticks. I'm sure you can get an "A" with very little effort. Good luck.
2006-09-21 19:32:16 · answer #3 · answered by arejokerswild 6 · 0⤊ 0⤋
It is really nice of these people to do your homework for you, but they are doing you a dis-service. Homework is an educational tool and you are defeating its purpose by asking others to do it for you.
2006-09-21 20:14:40 · answer #4 · answered by joker_32605 7 · 0⤊ 0⤋
find dy upon dx first,itz
12xsquare minus 2x. since gradient is zero,itz a horizontal line.when x=0,y too equals to zero....then?????
2006-09-21 01:43:53 · answer #5 · answered by sweetfloss8 2 · 0⤊ 1⤋