How do you find the equation tangent to #y=x^4-3x^2+2# at Point: (1,0)?

Question

Aurora Alvarado · Accepted Answer

#2x+y-2=0#.

We recall that, slope of the tgt. to the given curve at the pt. #(1,0)# is #[dy/dx]_(x=1,y=0)#. #y=x^4-3x^2+2 rArr dy/dx=4x^3-6x# # rArr [dy/dx]_(x=1,y=0)=4-6=-2#. Thus, the slope of the tgt. is #-2#, it passes thro. the pt. #(1,0)#. Hence, eqn. of the tgt. is# : y-0=-2(x-1), i.e., 2x+y-2=0#.

Emily Ammerman · Answer

undefined To find the equation of the tangent line to the curve y = x^4 - 3x^2 + 2 at the point (1,0), we need to find the slope of the tangent line at that point and then use the point-slope form of a linear equation.

To find the slope, we take the derivative of the given function with respect to x. The derivative of y = x^4 - 3x^2 + 2 is dy/dx = 4x^3 - 6x.

Next, we substitute x = 1 into the derivative to find the slope at the point (1,0). dy/dx = 4(1)^3 - 6(1) = -2.

So, the slope of the tangent line at (1,0) is -2.

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: y - 0 = -2(x - 1).

Simplifying the equation, we get y = -2x + 2.

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