If G(x) = x^2 − 3x + 3, find G'(a) and use it to find equations of the tangent lines to the curve y = x^2 − 3x + 3 at the points (0, 3) and (2,1).

Answer 1

I hope this is right. I don't have time to check it though, and no way to save it, so here it is:

G'(a)=2x-3 which is the slope of the tangent line

At the point (0,3) x=0, plug that into G'(a)=2x-3 to get -3 as the slope of the tangent line

For (2,1) plug 2 into G'(a)=2x-3 to get 2(2)-3=1 as the slope of the tangent line

Go back to the original equation G(x)=x^2-3x+3 and plug in the x values for the points

At (0,3) x=0 so G(0)=3. That's the y intercept. With the slope & y intercept use the slope intercept

form of the equation y=mx+b m=slope b=y intercept. y=-3x+3

For the other point, (2,1), G(2)=(2)^2 - 3(2) +3 = 4-6+3=1 That's the y intercept. 1 is also the slope

so y=x+1 is the equation of the tangent line to the quadratic equation at that point.

Answer 2

You can find the derivative using the Power rule. The tangent line is the line going thru the given point with slope equal to the derivative.