To find the values of a and b, we need to use the given information that the normal to the curve at x = 1 has a gradient of 1 and intercepts the y-axis at (0, -14).
We'll start by finding the gradient of the curve at x = 1. To do this, we'll differentiate the equation y = ax^(1/2) + bx with respect to x.
Differentiating y = ax^(1/2) + bx:
dy/dx = (1/2)ax^(-1/2) + b
Now, substitute x = 1 into the equation to find the gradient at x = 1:
dy/dx = (1/2)a(1)^(-1/2) + b
dy/dx = (1/2)a + b
Given that the gradient at x = 1 is 1, we have:
(1/2)a + b = 1 ...equation 1
Next, we'll find the equation of the line that represents the normal at x = 1. The equation of a line can be written as y = mx + c, where m is the gradient and c is the y-intercept.
Since the gradient of the normal is 1 (perpendicular to the gradient of the curve), the gradient of the line is -1 (negative reciprocal of 1).
At x = 1, the coordinates on the curve are (1, a(1/2) + b). We'll substitute these coordinates into the equation y = mx + c to find c (the y-intercept):
-14 = -1(1) + c
-14 + 1 = c
c = -13
Therefore, the equation of the line representing the normal at x = 1 is:
y = -x - 13 ...equation 2
Now, we'll find the point where the curve intersects with the line representing the normal at x = 1.
Setting the equations for the curve and the line equal to each other:
ax^(1/2) + bx = -x - 13
ax^(1/2) + bx + x + 13 = 0
Let's now substitute x = 1 into this equation:
a(1^(1/2)) + b(1) + 1 + 13 = 0
a + b + 14 = 0 ...equation 3
From equation 1, we have:
(1/2)a + b = 1
Let's now solve equations 1 and 3 simultaneously. Multiply equation 1 by 2 to eliminate the fraction:
a + 2b = 2 ...equation 4
Subtract equation 4 from equation 3:
(a + b + 14) - (a + 2b) = 0 - 2
a + b + 14 - a - 2b = -2
- b + 14 = -2
- b = -16
b = 16
Now, substitute this value of b into equation 1:
(1/2)a + 16 = 1
1/2a = 1 - 16
1/2a = -15
a = -15 * 2
a = -30
Therefore, the values of a and b are -30 and 16 respectively, yielding the equation of the curve as y = -30x^(1/2) + 16x.