Question 750300: Find equations of trhe tangents to the curve f(x) = x2 - 9x + 20 at points where the curve crosses the x-axis Answer by FrankM(1040) (Show Source):
You can put this solution on YOUR website! X^2-9x+20=0
(x-4)(x-5)=0
So the parabola crosses the xaxis at 4 and 5.
The slope at any point on the curve is 2x-9 (the derivative of the equation)
Edit - I'm not sure how to explain this without referencing introductory calculus. The slope of any point on a parabola aX^2+bX+c is the x value substituted into the equation 2aX+b. This is the first derivative of that general equation.
At (4,0) slope is -1 y=-x+4
At (5.0) slope is +1 y=x-5
