How do you find the equation of the line tangent to #f(x)=2x^2# at x=-1?

Question

Ezekiel Alexander · Accepted Answer

#y=-4x-2# #"we require the slope m and a point on the line to"# #"obtain the equation of tangent"# #•color(white)(x)m_(color(red)"tangent")=f'(-1)# #rArrf'(x)=4xrArrf'(-1)=-4# #f(-1)=2(-1)^2=2rArr(-1,2)" is point on line"# #rArry-2=-4(x+1)# #rArry=-4x-2larrcolor(red)"equation of tangent"#

Ezekiel Alexander · Answer

undefined To find the equation of the line tangent to f(x)=2x^2 at x=-1, we need to find the slope of the tangent line and the point of tangency.

First, we find the derivative of f(x) with respect to x, which gives us f'(x)=4x.

Next, we substitute x=-1 into f'(x) to find the slope of the tangent line at x=-1. So, f'(-1)=4(-1)=-4.

Now, we have the slope of the tangent line, which is -4.

To find the point of tangency, we substitute x=-1 into f(x). So, f(-1)=2(-1)^2=2.

Therefore, the point of tangency is (-1, 2).

Finally, we can use the point-slope form of a linear equation, y-y1=m(x-x1), where m is the slope and (x1, y1) is a point on the line.

Substituting the values, we have y-2=-4(x-(-1)).

Simplifying the equation, we get y-2=-4(x+1).

Expanding further, we have y-2=-4x-4.

Rearranging the equation, we get y=-4x-2.

Thus, the equation of the line tangent to f(x)=2x^2 at x=-1 is y=-4x-2.