How do you find the equation of the line tangent to the graph of #y=3x^2-6x+2#, at the point (3,11)?

Question

William Abdalla · Accepted Answer

#y-12x+25=0#

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

Apply differentiation (power rule )

#y'=6x-6#

substitute with #x# value to get the slope #m#

#y'=12#

substitute in the equation of the tangent with the point and slope

#y-y_1=m(x-x_1)#

where #(x_1,y_1)=(3,11)#

#y-11=12(x-3)#

Simplify

#y-12x+25=0#

I hope I helped.

Cameron Alexander · Answer

undefined To find the equation of the line tangent to the graph of y=3x^2-6x+2 at the point (3,11), we need to find the slope of the tangent line at that point.

First, we find the derivative of the given function y=3x^2-6x+2. The derivative of y with respect to x is dy/dx = 6x - 6.

Next, we substitute x=3 into the derivative to find the slope at the point (3,11). dy/dx = 6(3) - 6 = 12.

Therefore, the slope of the tangent line at the point (3,11) is 12.

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 substitute the values (3,11) and 12 into the equation.

y - 11 = 12(x - 3)

Simplifying the equation, we get:

y - 11 = 12x - 36

Rearranging the equation to the standard form, we have:

12x - y = 25

Therefore, the equation of the line tangent to the graph of y=3x^2-6x+2 at the point (3,11) is 12x - y = 25.