Help i'm confused on what to do next

consider the function f(x)=x(e^x+16). Find the equation of the line tangent to f(x) at x=0.

i got the derivate but, what do i do next? Thanks

dy/dx=(e^(x)*(x+1)+16)
this can also be written as f'(x)=(e^(x)*(x+1)+16)|x=0

f'(0)=(e^(0)*(0+1)+16)
=1*1+16=17.

The derivative at x=0 is the instantenous rate of change at that point, also known as the slope.

Warning: comical algebra throwback ahead...

remember that seemingly annoying equations y-yo=m(x-xo)

You'll become good friends again if you've neglected him in the past few courses.

evaluate f(0) again; 0(e^(0)+16) = 0, your (Xo,Yo)=(0,0)
thus,

y-0=17(x)
y=17x

Hope this helps you, if not let me know.

they want the equation of a line that is tangent to f(x) at x=0. With the derivative you have the slope of the line. They gave you the point at which that want that line to pass through at x= 0, which is (0, f(0))=(0,0).

The eqn of a line is y=mx+b. it goes through (0,), so b = 0;
y=mx

y = (derivative of f(x) at 0) * x = f'(0) * x
f'(x) = (e^x +16) + xe^x
f'(0)= (1+16) + 0 = 17
y= 17x