To find the tangent line you need to know the gradient (m) of the tangent and a point of the tangent (x1,y1) to eventually fill the form y-y1=m(x-x1)
the first part of the question asked you to find the derivative of the initial equation, so if you were to find f'(-3) youd also find the gradient to the tangent line at that point, so:
f'(-3)=7/2sqrt(7(-3)+25)
=7/2sqrt(4)
=7/4
next we need the point (x1, y1), we know x1=-3 so all we do is plug that into f(x)
f(-3)=sqrt(7(-3)+25)
=sqrt(4)
=2
therefore (x1, y1) = (-3,2)
so plug into the equation for a straight line and:
y-2=7/4 (x-(-3))
y-2=7/4 (x+3)
y-2=7x/4 + 21/4
y=7x/4 +29/4
hope this made sense, good luck
