Question

Прямая y=-7x-9 является касательной к графику функции f(x)=4x^2+bx. Найдите b, учитывая, что абсцисса точки касания больше 0.Помогите, пожалуйста, с подробным решением!

Answer 1

Если прямая является касательной,то f`(x0)=-7
f`(x)=8x+b
8x+b=-7
8x=-b-7
x=(-b-7)/8
(-b-7)/8>0
-b-7>0
b<-7
b∈(-∞;-7)

Answer 2

Problem Analysis

We are given that the equation of a line, y = -7x - 9, is a tangent to the graph of the function f(x) = 4x^2 + bx. We need to find the value of b, given that the x-coordinate of the point of tangency is greater than 0.

Solution

To find the value of b, we need to determine the derivative of the function f(x) and equate it to the slope of the tangent line.

The derivative of f(x) = 4x^2 + bx with respect to x is given by: f'(x) = 8x + b

Since the line y = -7x - 9 is a tangent to the graph of f(x), the slope of the tangent line is equal to the derivative of f(x) at the point of tangency.

Therefore, we have: -7 = 8x + b

To find the x-coordinate of the point of tangency, we need to solve the equation f(x) = -7x - 9. Substituting f(x) = 4x^2 + bx, we get: 4x^2 + bx = -7x - 9

Simplifying the equation, we have: 4x^2 + (b + 7)x + 9 = 0

Since the x-coordinate of the point of tangency is greater than 0, we can assume that there is only one solution to the equation. Therefore, the discriminant of the quadratic equation should be equal to 0.

The discriminant, D, is given by: D = (b + 7)^2 - 4(4)(9)

Setting D = 0 and solving for b, we have: (b + 7)^2 - 144 = 0 (b + 7)^2 = 144 b + 7 = ±√144 b + 7 = ±12

Solving for b, we have two possible values: b = -7 + 12 = 5 b = -7 - 12 = -19

However, since the x-coordinate of the point of tangency is greater than 0, we can eliminate the solution b = -19.

Therefore, the value of b is 5.

Conclusion

The value of b, given that the equation y = -7x - 9 is a tangent to the graph of f(x) = 4x^2 + bx and the x-coordinate of the point of tangency is greater than 0, is 5.