As shown in the figure, the positive ratio function image passes through point a, and the analytic expression of the function is? Point (1,3) a is connected with point (0,0) as a straight line

Let the analytic expression of the positive proportional function be y = KX
Substituting point (1,3) into the analytic expression, we get 3 = K,
So the analytic expression of the function is y = 3x

Given that the image of a positive ratio function passes through point (2,3), find the function

Positive ratio function y = kx
So 3 = 2K, k = 3 / 2
So y = (3 / 2) X

If the image of a positive ratio function passes through (2, - 3), the analytic expression is: I need a procedure,

Because the function is proportional,
So let y = KX, substitute (2, - 3) into - 3 = 2K, k = - 2 / 3
So the function is y = - 2 / 3x
The image of y = -- 2x + 6 passes through the first, second and fourth quadrants

RELATED INFORMATIONS

1. It is known that the image of positive ratio function passes (- 4,8) 1) if the points (a, - 1) and (2, - b) are on the image, the values of a and B are obtained
2. It is known that Y-3 is in positive proportion to x, and when x = 2, y = 7. (1) translate the function image to pass through the point (2. - 1) to find the analytical expression of the translated line
3. The function y = 2Sin (3x + π / 6) image is shifted to the left by π / 6 units, and the function y = 2cos (3x + π / 6) image is obtained. Please explain whether this conclusion is correct
4. The image of function y = 2Sin (2x + π / 6) can be obtained by which unit the image of function y = 2sin2x is shifted
5. The area of the triangle formed by the line y = KX + 3 and the two coordinate axes is 9, and the value of K is obtained
6. It is known that the line y = KX + B passes through the point (2.5,0), and the area of the triangle surrounded by the coordinate axis is 6.25. The function analytical formula of the line is obtained
7. The image of the positive ratio function y = x and the image of the inverse ratio function y = x (K ≠ 0) intersect with point a at the first term limit, and make a vertical line of X axis through point a, with the perpendicular foot M
8. The image of positive scale function passes through points (2, - 4) and makes a vertical line of X axis through a point a on the image. The vertical foot is B (4,0), Finding the coordinates of point a and the area of triangle AOB
9. The image of a positive scale function passes through a point (- 2,8) and passes through a point a on the image to form a vertical line of y-axis. The vertical foot is B (0,6)
10. If the area of the triangle formed by the line y = 2x + B and the coordinate axis is 6, then the analytic expression of the function of the line is
11. Let f (x) = x-x / 1-alnx, if f (x) has two extreme points x1x2, X1 belongs to the range of (0,1) x2 and (1,2) a
12. It is known that f (x) is a decreasing function on a set of real numbers. If a + B ≤ 0, then the following is true? A.f（a）+f（b）≤-{f（a）+f（b）} B.f（a）+f（b）≤f(-a)+f(-b） c.f(a）+f（b）≥-｛f（a）+f（b）｝ d.f（a）+f（b）≥f（-a）+f（-b） 2. Given that f (x) is an odd function, if x > 0, f (x) = X-1, then f (x)
13. The known function f (x) = ((x ^ 2) / 2) - alnx (a)
14. Given the function f (x) = x2 + (2a-1) x-alnx, G (x) = - 4 / x-alnx, (a ∈ R) The function f (x) = x2 + (2a-1) x-alnx, G (x) = - 4 / x-alnx (a ∈ R) is known (1) When a ＜ 0, find the minimum of F (x); (2) If the image of the function y = f (x) and y = g (x) has two different intersections m, N on X ∈ [1,3], the value range of a is obtained
15. Given the function f (x) = alnx / (x + 1) + B / x, the tangent equation of curve y = f (x) at point (1, f (1)) is x + 2y-3 = 0 (1) Finding the value of a and B (2) If x ＞ 0 and X ≠ 1, f (x) ＞ LNX / (x-1) + K / x, find the value range of K I have solved the first problem, a = b = 1. The second problem uses partial derivation, let H (x) = 1 / (1-x ^ 2) [2lnx + (k-1) (x ^ 2-1) / x], and then let g (x) = 2lnx + (k-1) (x ^ 2-1) / X to solve the derivation. What should I do next?
16. Given the function f (x) = 2 / x + alnx, a ∈ R (1) If a = 4, find the monotone interval of function f (x) (2) If the function f (x) increases monotonically on [1, + 00], find the value range of real number a
17. The known function f (x) = x ^ 2 + alnx
18. Given the function f (x) = alnx - (1 + a) x + 0.5x ^ 2, (a belongs to R), if f (x) > = 0, it holds for any X in the domain of definition, and the value range of a is obtained
19. Let f (x) = 1 (x = 0) LG | x | (x ≠ 0) defined on R. if the equation F2 (x) + BF (x) + C = 0 has exactly three different real solutions x1, X2, X3, then X12 + X22 + X32=______ ．
20. If the function f (x) = 2A & # 178; + (A-1) x + 3 is even, then a=