What is the slope-intercept equation of the line tangent to the graph of y=8ln(x) when x=5.
To find the slope-intercept equation of the line tangent to the graph of y=8ln(x) at a specific point, you need to calculate both the slope and the y-intercept.
The slope of the tangent line is equal to the derivative of the function y=8ln(x) at the given point. In this case, we want to find the slope of the tangent line when x=5. To do this, we need to take the derivative of y=8ln(x) with respect to x.
The derivative of ln(x) with respect to x is 1/x. Therefore, the derivative of 8ln(x) with respect to x is 8/x.
Substituting x=5 into the derivative expression gives us the slope of the tangent line at the point x=5:
slope = 8/5
Now that we have the slope, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by:
y - y₁ = m(x - x₁)
where (x₁, y₁) represents the coordinates of the point on the line, and m represents the slope.
In this case, we know that the tangent line passes through the point (5, 8ln(5)). Substituting these values into the point-slope form, we get:
y - 8ln(5) = (8/5)(x - 5)
To simplify this equation, you can distribute the (8/5) to get:
y - 8ln(5) = (8/5)x - 8
Finally, you can rearrange the equation to the slope-intercept form, which is in the form y = mx + b, where m represents the slope and b represents the y-intercept:
y = (8/5)x - 8 + 8ln(5)
Therefore, the slope-intercept equation of the line tangent to the graph of y=8ln(x) when x=5 is:
y = (8/5)x - 8 + 8ln(5)