How do you find the equation of the line tangent to #y= -x^3+6x^2-5x# at (1,0)?

Question

Charles Anctil · Accepted Answer

#y=4x-4#

To find the tangent we will use the equation #(y-y_1)=m(x-x_1)#,

where #(x_1,y_1)# is the point at which we are finding the tangent, and #m# is the gradient.

Since we know #(x_1,y_1)# we have to find the gradient of the tangent, this is done by differentiation.

#y=-x^3+6x^2-5x#

#(dy)/dx=-3x^2+12x-5#

#m=[(dy)/dx]_(x=1)=-3xx1^2+12xx1-5#

#m=[(dy)/dx]_(x=1)=-3+12-5=4#

eqn of tangent

#(y-0)=4(x-1)#

#=>y=4x-4#

Elijah Abram · Answer

undefined To find the equation of the line tangent to the curve y = -x^3 + 6x^2 - 5x at the point (1,0), we need to find the slope of the tangent line at that point.

First, we find the derivative of the given function y = -x^3 + 6x^2 - 5x.

The derivative of y with respect to x is dy/dx = -3x^2 + 12x - 5.

Next, we substitute x = 1 into the derivative to find the slope at the point (1,0).

dy/dx = -3(1)^2 + 12(1) - 5 = 4.

Therefore, the slope of the tangent line at (1,0) is 4.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

Plugging in the values, we have y - 0 = 4(x - 1).

Simplifying, we get y = 4x - 4.

Thus, the equation of the line tangent to y = -x^3 + 6x^2 - 5x at (1,0) is y = 4x - 4.