Okay so I found the derivative of f(x) which is f'(x) = -2x+3. Then I set it equal to the function y and found the point of intersection. So that was the answer for a as I'm guessing
Then for b I used that Point of intersection and the perpendicular slope of 1/3x+2 which is -3 and found the equation of the line. I feel I'm doing something wrong though...
Hopefully Chris is doing the math and comes up with the same answer I calculated...here is my graph
Ok -2x +3 gives the SLOPE of the parabola at any point x
We want the slope to be -3 (negative inverse of 1/3.....the slope of the line)
so -2x+3 = -3 yields x = 2 is where the tangent line is slope -3
Sustitute x= 2 into the ORIGINAL equation to find y = 2
Now find the line given slope = -3 and point 3,2
2= -3(3 ) + b so b = 11
y = -3x + 11 (see graph)
The question is...
What are the coordinates of the point on the graph of f(x) where the slope of f(x) = -3 ?
the slope of f(x) at any x value = f'(x) = -2x + 3
What x value makes the slope of f(x) be -3 ?
What x value makes f'(x) be -3 ?
f'(x) = -3
-2x + 3 = -3
-2x = -6
x = 3
When x = 3 , the slope of f(x) = -3
When x = 3 , f(x) = f(3) = 2
So at the point (3, 2) , the slope of f(x) is -3 .