How do you find the equation of the line tangent to #f(x) = x^2 + 2x +1#, with the point (-3, 4)?

The equation of the tangent is : y - b = m(x - a ) where m

represents the gradient and (a , b ) a point on the line.

Require to find m , (a , b ) is given (-3 , 4 ) Now m = f'(x).

f'(x) = 2x + 2 and f'(-3) = 2(-3) + 2 = - 6 + 2 = - 4 = m

Equation is : y - 4 = -4 ( x + 3 )

ie y - 4 = - 4x - 12

To find the equation of the line tangent to the function f(x) = x^2 + 2x + 1 at the point (-3, 4), we need to find the derivative of the function and evaluate it at x = -3.

The derivative of f(x) = x^2 + 2x + 1 is f'(x) = 2x + 2.

To find the slope of the tangent line, we substitute x = -3 into the derivative: f'(-3) = 2(-3) + 2 = -4.

The slope of the tangent line 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 substitute (-3, 4) and -4 into the equation:

y - 4 = -4(x - (-3)).

Simplifying, we get the equation of the tangent line: y - 4 = -4(x + 3).

To find the equation of the tangent line to the function ( f(x) = x^2 + 2x + 1 ) at the point ((-3, 4)), follow these steps:

1. Find the derivative of the function ( f(x) ) using the power rule. The derivative of ( f(x) ) is ( f'(x) = 2x + 2 ).

2. Evaluate the derivative at the x-coordinate of the given point (-3) to find the slope of the tangent line. ( f'(-3) = 2(-3) + 2 = -4 ).

3. Use the point-slope form of the equation of a line to write the equation of the tangent line. The point-slope form is ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is the given point and ( m ) is the slope.

4. Substitute the values of the given point and slope into the point-slope form. ( y - 4 = -4(x + 3) ).

5. Simplify the equation to obtain the equation of the tangent line. ( y - 4 = -4x - 12 ). This simplifies to ( y = -4x - 8 ).

Thus, the equation of the tangent line to ( f(x) = x^2 + 2x + 1 ) at the point ((-3, 4)) is ( y = -4x - 8 ).