Math derivative tangent line question ?

Question

the problem is

find equation of tangent line to the graph of y = f(x) at the point where x = -3 if f(-3) = 2 and f'(-3) = 5

im new with derivative so can u ploease show me the steps thanks

Answer 1

the derivative gives the slope of the function (and therefore the slope of a tangent line at any given point)

since f ' (-3) = 5, that means the slope of the tangent will be 5 at x = 3

since f(-3) = 2, that means the intersection of the function and tangent will be the point (-3 , 2)

now you have a slope and a point; you can write the equation from there...

y = mx + b

y = 5x + b

but you know that (-3 , 2) is on the line, so you can substitute in to find b:

2 = 5(-3) + b

2 = -15 + b

b = 17

so the equation of the tangent line is y = 5x + 17

(or in standard form: 5x - y = -17)

Answer 2

The tangent line at f(a) is a line (in the form y = mx+b) with the slope f'(a). In this case the slope of the tangent line to f(x) at f(-3) is f'(-3) = 5. Now you have the slope of a line and a point on the line, f(-3), so you can solve for the equation of the line.

y = mx + b

y = 5x + b   **m = f'(-3) = 5, y = f(-3) = 2 (both from question)

2 = 5(-3) + b

17 = b

Therefore, the equation of the tangent line to f(x) at x = -3 is...

y = 5x + 17.

Answer 3

Hey,

this a very straight forward problem

the point slope formula is y-b=m(x-a), this a formula for a line that passes through point (a,b) with a slope of m.

So you have f(-3)=2 which says that the tangent will pass through point (-3,2).

if you plug into the formula you get y-2=m(x+3) and now all you need is the slope of the line and thats given because f'(-3)=5 represents the slope of the tangent line when x=3

so you have y-2=5(x+3) solve for y and you have your equation of the line

hope this helps