Question

Normal and tangent lines at a point.how to solve the tangent,normal line and slope of normal line?

Answer 1

y=x^3-5x+4 at (1,0)

Answer 2

derive y first to get the slope of the tangent line
y'=3x^2-5
Use point 1,0 in (y-0)=y'(x-1)

Normal at 1,0 uses -(y'(y-0))=(x-1)

Answer 3

how to get the tangent?

Answer 4

slope of normal i believe is (-1/y')

Answer 5

am how did you get it?pls explain

Answer 6

are you doing calculus? If so, derivative determines the slope of the tangent line at that point. If so, you can plug in the derivative to the equation (derivative)*(X-x)=(Y-y) for the tangent line. Small x is given as 1 and small y is given as 0. The slope of the normal as given in pre-calculus is the opposite of the reciprocal. For example if the slope was 1 then the normal would be -1/1 and this would be similar to the derivative of y. -1/y would be the slope of the normal.

Answer 7

you would plug the -1/deriviative in the equation i gave in the last post and not the first one.