Question

Let f(x)=4 x+9-3 eˣ. Then the equation of the tangent line to the graph of f(x) at the point (0,6) is given by y=m x+b for

[ m=; b= ]

Answer 1

The equation of the tangent line to the graph of f(x) = 4x + 9 - 3e^x at the point (0,6) is y = x + 6.

Step-by-step explanation:

The equation of the tangent line to the graph of the function f(x) = 4x + 9 - 3e^x at the point (0,6) can be found using the derivative of the function.

First, take the derivative of f(x) with respect to x. The derivative of 4x is 4, and the derivative of 3e^x is 3e^x. Therefore, the derivative of f(x) is 4 - 3e^x.

Next, substitute x=0 into the derivative to find the slope of the tangent line at the point (0,6).

y=4-3e^0=4-3=1

So, the slope of the tangent line is 1.

Since the tangent line passes through the point (0,6), we can use the point-slope form of the equation of a line to find the equation of the tangent line.

y - y1 = m(x - x1), where (x1, y1) is the point (0,6) and m is the slope.

Substituting the values, we get y - 6 = 1(x - 0), which simplifies to y - 6 = x.

Therefore, the equation of the tangent line is y = x + 6.