Question

Suppose for a function y = f(x) that f(-2) = 3 and f'(-2) = 23 . Write the equation of the tangent line to f(x) at x = -2.

Answer 1

Answer:

The equation of the line is $y=23x+49$

Step-by-step explanation:

The general equation of a line is given by

$y=mx+c$

Where,

'm' is the slope of the line

'c' is the intercept of the line

Now we are given that $f'(x)|_{x=-2}=23$

By definition the derivative of a function at any point is the slope of the tangent at that point

Thus the slope of the line passing x = -2 is 23

Hence 'm' = 23

Thus the equation of the line becomes

$y=23x+c$

to find the intercept we are given that the line passes through (-2,3)

Using this information in the above equation of line we get

$3=23\times -2+c\\\\\therefore c=49\\\\\therefore y=23x+49$

Answer 2

Answer:

The equation of tangent line is $y=23x+49$.

Step-by-step explanation:

Given information: y = f(x) that f(-2) = 3 and f'(-2) = 23 .

The given function is

$y=f(x)$

Differentiate with respect to x.

$y'=f'(x)$

We need to find the equation of the tangent line to f(x) at x = -2.

Slope of tangent line is

$y'_{[x=-2]}=f'(-2)=23$

Slope of line is 23.

At x=-2 the value of function is 3. It means the tangent line passes through the point (-2,3).

Equation of tangent line is

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

$y-3=23(x-(-2))$

$y-3=23x+46$

Add 3 on both sides.

$y=23x+49$

Therefore the equation of tangent line is $y=23x+49$.