A function y= ex + f + dx^2 intercepts on the y-axis at point (0,-1). Determine what d, e, and f are when points (1,2) and (-1, -2) lie on the curve.
y= ex + f + dx²
When (0,-1)
ex+f+dx²=y
0+f+0=-1
F=-1
at(1,2)
ex+f+dx²=y
e+f+d=2
e+d=3.......(1)
At(-1, -2)
ex+f+dx²=y
-e+f+d=-2
-e+d=-2+1
-e+d=-1.....(2)
-------------------------------------------------
e+d=3.......(1)
-e+d=-1.....(2)
2d=2
d=1
e=2
I had e,f,d=(2,-1,1)
(0,-1) lies on y= ex + f + dx^2
-1 = 0 + f + 0
f = -1
so y = ex -1 + dx^2
(1,2) lies on y= ex -1 + dx^2 ----> 2 = e -1 + d or e + d = 3
(-1,-2) lies on y= ex -1 + dx^2 --> -2 = -e -1 + d or -e + d = -1
add them :
2d = 2
d = 1
back in 2 = e - 1 + d
2 = e -1 + 1
e = 2
So you have
y = 2x -1 + x^2
To determine the values of d, e, and f in the function y = ex + f + dx^2, we can use the given information about the intercept on the y-axis and the points on the curve.
Let's start with the intercept on the y-axis. When x = 0, the function intersects the y-axis at point (0, -1). We can substitute these values into the equation:
-1 = e(0) + f + d(0)^2
-1 = f
So we have found that f = -1.
Now let's use the given points (1, 2) and (-1, -2) to determine the values of e and d. Substituting these points into the equation, we get:
2 = e(1) - 1 + d(1)^2
-2 = e(-1) - 1 + d(-1)^2
Simplifying these equations, we have:
2 = e - 1 + d
-2 = -e - 1 + d
Adding these two equations together eliminates the e term:
2 + (-2) = (e - 1) + (-e - 1) + d + d
0 = -2
Since 0 = -2 is not a true equation, it means that there is no solution that satisfies both equations simultaneously. Therefore, the given function y = ex + f + dx^2 does not pass through the points (1, 2) and (-1, -2).