Find the equation of the tangent line to the graph of y=3^x−2 at x = 1. Give your answer in slope-intercept form. Show all your work.
To find the equation of the tangent line to the graph of y=3^x-2 at x = 1, we need to find the slope and a point on the tangent line.
First, let's find the slope. The slope of the tangent line is given by the derivative of the function at x = 1. Let's find the derivative of y=3^x-2.
Using the power rule of differentiation, the derivative of 3^x is ln(3) * 3^x. The derivative of -2 is 0, as it is a constant.
So, the derivative of y=3^x-2 is dy/dx = ln(3) * 3^x.
Now let's find the slope at x = 1. Substituting x = 1 into the derivative, we get dy/dx = ln(3) * 3^1 = ln(3) * 3.
So, the slope of the tangent line at x = 1 is ln(3) * 3.
Next, let's find a point on the tangent line. We have the point (1, y=3^1-2) = (1, 3-2) = (1, 1).
Now we can use the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept, to find the equation of the tangent line.
Using the values we found, the equation of the tangent line is y = (ln(3) * 3)x + b.
To find b, we substitute the point (1, 1) into the equation:
1 = (ln(3) * 3)(1) + b.
Simplifying, we get 1 = 3ln(3) + b.
Subtracting 3ln(3) from both sides, we get 1 - 3ln(3) = b.
So, the equation of the tangent line in slope-intercept form is y = (ln(3) * 3)x + 1 - 3ln(3).