# Find an equation for the tangent line to the given curve at the point where x=x0. y=(x^2+3x-1)(2-x); x0=1

#1

Find an equation for the tangent line to the given curve at the point where x=x0.

y=(x^2+3x-1)(2-x); x0=1

for a tangent line the slope is the derivative which is defined as limh→0(f(x+h)−f(x))/h
another helpful rule if youre comfortable with derivatives is that
Dx(f(x)*g(x)) = f’(x)g(x)+g’(x)f(x)

that being said, if f(x)=(x^2+3x-1) and g(x)=(2-x)
y=f(x)g(x)
f’(x)=2x+3
g’(x)=-1
so y’ = -(x^2+3x-1)+((2x+3)(2-x))=slope at x
that simplifies to -3x^2 -2x +7
and it specifies x=1 so the value for slope is 2

ok now we need an x and y coordinate.

since they give x=1 and we know a tangent touches the function at one point we just need the y value of the original function at x=1.
y=(1+3-1)(2-1) at x=1
=3*1=3
so x=1
y=3
slope=2
just use y-y1=m(x-x1)
y-3=2(x-1)

and thats the equation you’re looking for!

feel free to manipulate it algebraically to put it into y=mx+b if you want by solving for y