The straight line y = 9x + 5 is tangent to the graph of the function 18x ^ 2 + bx + 7.

The straight line y = 9x + 5 is tangent to the graph of the function 18x ^ 2 + bx + 7. Find b, given that the abscissa of the touch point is less than 0.

Let x0 be the required abscissa of the tangency point.

Then the derivative of the function y = 18x ^ 2 + bx + 7 at the point x0 must be equal to 9.

Find the derivative of the function y = 18x ^ 2 + bx + 7:

y ‘= (18x ^ 2 + bx + 7)’ = 2 * 18x + b = 36x + b.

Therefore, the ratio must be satisfied:

36×0 + b = 9,

whence it follows that b = 9 – 36×0.

Since the straight line y = 9x + 5 touches the graph of the function y = 18x ^ 2 + bx + 7 at the point x0, the following relation must be fulfilled:

9×0 + 5 = 18×0 ^ 2 + bx0 + 7.

Substituting the found value b = 9 – 36×0 into this ratio, we get:

9×0 + 5 = 18×0 ^ 2 + (9 – 36×0) * x0 + 7;

9×0 + 5 = 18×0 ^ 2 + 9×0 – 36×0 ^ 2 + 7;

9×0 + 5 = 9×0 – 18×0 ^ 2 + 7;

5 = -18×0 ^ 2 + 7;

18×0 ^ 2 = 7 – 5;

18×0 ^ 2 = 2;

x0 ^ 2 = 2/18;

x0 ^ 2 = 1/9;

x0 ^ 2 = (1/3) ^ 2;

x0_1 = 1/3;

x0_2 = -1/3.

According to the condition of the problem, the abscissa of the touching point must be less than 0, therefore, the value x0 = 1/3 is not suitable.

Knowing x0, we find b:

b = 9 – 36×0 = 9 – 36 * (-1/3) = 9 + 12 = 21.

Answer: b = 21.