If the image of the function y = x ^ 2-3x + 2 is translated up and down along the Y axis and passes through the point (3,1), then the direction and distance of translation are

If the image of the function y = x ^ 2-3x + 2 is translated up and down along the Y axis and passes through the point (3,1), then the direction and distance of translation are

Y = x ^ 2-3x + 2 moves up and down along the y-axis, then x will not move. When x = 3, y = 2, then the original direction of y = x ^ 2-3x + 1 moves up 1 unit

The solution of 3x-3 = y + 5, 5y-5 = 3x + 15

3x-3=y+5
5y-5=3x+15
The result is: 3x-y = 8
3x-5y=-20
Up and down: 4Y = 28
y=7
Then: x = 5

3X-Y=5 5Y-1=3X＋5
The title is given,
The final answer is--
X = 31 / 12
Y = 11 / 4,

Let 3x-y = 5 be Formula 1
5y-1 = 3x + 5 is formula 2
① Formula + formula 2: 3x-y + 5y-1 = 5 + 3x + 5
That is: 4y-1 = 10, then y = 11 / 4
If y = 11 / 4 is taken into Formula 1, 3x-11 / 4 = 5
Then x = 31 / 12

The known function f (x) = x & # 179; + ax & # 178;, a

(2011. Qionghai the first mock exam) parabola y=4x2 the shortest distance from the last point to the straight line y=4x-5 is the coordinates of that point.
A. （1，2）B. （0，0）C. (12，1)D. （1，4）

Y '= 8x, x = 12 is obtained from 8x = 4, so the tangent point coordinate of the tangent with the slope of 4 of the parabola is (12,1), and the distance from the point to the straight line y = 4x-5 is the shortest

Find a point P on the parabola y'2 = 4x so that the distance from P to the straight line X-Y + 4 = 0 is the shortest

The distance between curve and straight line is the shortest
The results of simultaneous straight line parabola are as follows:
(x+k)^2-4x=0
That is, x ^ 2 + (2k-4) x + K ^ 2 = 0
△=(2k-4)^2-4k^2=0
We get k = 1
X ^ 2-2x + 1 = 0, x = 1, then y = 2
If the line X-Y + 1 = 0 is tangent to the parabola, then the coordinates of point P are (1,2)
Then the shortest distance d = (4-1) / √ (2) = 3 √ 2 / 2

Find a point P on the square of the parabola y = 4x, so that the distance from point P to the straight line y = 4x-5 is the shortest

Let P abscissa be a, y = 4x ^ 2
So the ordinate 4A ^ 2
So the distance from P to 4x-y-5 = 0 = | 4a-a ^ 2-5 | / radical (4 ^ 1 + 1 ^ 2)
=|A ^ 2-4a + 5 | / radical 17
The shortest distance is the smallest molecule
|a^2-4a+5|=|(a-2)^2+1|
So when a = 2, the molecule is the smallest and the distance is the shortest
4a^2=16
So p (2,16)

If the point P is a point on the parabola y ^ 2 = 4x, then the coordinate of the point P with the shortest distance to the straight line y = x + 3 is?

X-Y + 3 = 0, let P point coordinate (y ^ 2 / 4, y), then the distance from P to the straight line d = (y ^ 2 / 4-y + 3) / root 2, find the minimum

The function f (x) = x + A / x + B (x is not equal to 0). If the tangent equation of the curve y = f (x) at point P (2, f (2)) is y = 3x + 1, find the analytic expression of F (x)
It's a process!

f’（x）=1-a/x^2
Because the tangent equation of curve y = f (x) at point P (2, f (2)) is y = 3x + 1
So f '(2) = 3,1-a / 4 = 3, a = - 8
Point (2,7) on f (x) image, 2-4 + B = 7, B = 9
So the analytic expression of F (x) is f (x) = X-8 / x + 9

1. Find the n-order derivative of function y = ln (x ^ 2 + 3x + 2), dny / DX ^ n. you must get the answer tomorrow morning!
2. Discuss the concavity and convexity and inflection point of function f (x) = x ^ 4-2x ^ 3 + 1

y = ln(x²+3x+2)dy/dx = (2x+3)/(x²+3x+2) = (2x+3)/[(x+2)(x+1)] = (x+2)^(-1) + (x+1)^(-1),n = 0d²y/dx² = (-1)(1!)(x+2)^(-2) + (-1)(1!)(x+1)^(-2),n = 1d³y/dx³ = (-1)²(2...