some help with this question would be much appreciated.
A curve y = f (x) has derivative f'(x) = 2x + 3 and intersects the line 5y - 18x - 64 = 0 at the points (2,20) and (-3,10). Find the area enclosed by the curve f(x) and the line.
some help with this question would be much appreciated.
A curve y = f (x) has derivative f'(x) = 2x + 3 and intersects the line 5y - 18x - 64 = 0 at the points (2,20) and (-3,10). Find the area enclosed by the curve f(x) and the line.
If y'=2x+3 then,
y=x^2+3x+C, for some C ----> Anti-derivative
The problem says the curve (parabola in this case) intersects at (2,20) which means when x=2 then y=20:
20=4+6+C thus, C=10.
The unique curve must be,
y=x^2+3x+10
The line is,
y=3.6x+12.8
The problem is that these two curve intersect in one other point which is not (-3,10) as you say. Thus there is no such curve.
Hello, 24680!
There are typos in this problem. .As written, it is contradictory.
Please check the original wording.
A curve y = f(x) has derivative f '(x) = 2x + 3
and intersects the line 5y - 18x - 64 = 0 at the points (2,20) and (-3,10).
Find the area enclosed by the curve f(x) and the line.
Contradiction . The point (-3,10) is not on the line 5y - 18x - 64 = 0
I suspect that you gave the wrong equation for the line.
The line through (2,20) and (-3,10) is: .2x - y + 16 .= .0
The curve turns out to be the parabola: .y .= .x² + 3x + 10
Am I close?