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

Question

Scarlett Allison · Accepted Answer

Equation of tangent is #4x+y+2=0#

At #x=1# ,#f(x)=2xx(-1)^2=2# hence tangent passes through #(-1,2)#

Slope of the line is given by value of derivative and as

#f'(x)=4x#, slope of tangent at #(-1,2)# will be #4xx(-1)=-4#

Hence equation of tangent is

#(y-2)=-4(x+1)# or

#4x+y+2=0#

Evelyn Agard · Answer

undefined To find the equation of the tangent line to the curve f(x) = 2x^2 at x = -1, we can follow these steps:

1. Find the derivative of the function f(x) with respect to x.
2. Evaluate the derivative at x = -1 to find the slope of the tangent line.
3. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope, to find the equation of the tangent line.

Let's go through these steps:

1. The derivative of f(x) = 2x^2 can be found using the power rule for differentiation. The power rule states that if f(x) = ax^n, then f'(x) = nax^(n-1). Applying this rule, we get f'(x) = 4x.

2. Evaluating the derivative at x = -1, we have f'(-1) = 4(-1) = -4. This is the slope of the tangent line.

3. Now, we can use the point-slope form of a line. Since the point of tangency is (-1, f(-1)), we need to find the corresponding y-coordinate. Plugging x = -1 into the original function, we get f(-1) = 2(-1)^2 = 2.

Using the point-slope form with the slope (-4) and the point (-1, 2), we have y - 2 = -4(x - (-1)).

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

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

Finally, rearranging the equation, we get the equation of the tangent line to the curve f(x) = 2x^2 at x = -1 as y = -4x - 2.