**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**

**Answer:**

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