Final answer:
To find the equation of the tangent line to the graph of the function f(x)=2e^(2x^2) + 3x at x=0, we can find the derivative of the function to get the slope of the tangent line. Plugging the values into the point-slope form of the equation of a line gives us the equation of the tangent line as y = 3x + 2.
Step-by-step explanation:
To find the equation of the tangent line to the graph of the function f(x)=2e^(2x^2) + 3x at x=0, we can first find the derivative of the function. The derivative of f(x) is given by f'(x) = 4x(2e^(2x^2)) + 3. Evaluating f'(x) at x=0 gives f'(0) = 3.
Since the tangent line has the same slope as the function at x=0, the slope of the tangent line is 3. We can use the point-slope form of the equation of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, to find the equation of the tangent line.
Plugging in the values x1 = 0, y1 = f(0) = 2e^(2(0)^2) + 3(0) = 2, and m = 3 into the equation, we get y - 2 = 3(x - 0), which simplifies to y - 2 = 3x. Therefore, the equation of the tangent line is y = 3x + 2.