If the instantaneous rate of change of f(x) at (4, −5) is 7, write the equation of the line tangent to the graph of f(x) at x = 4. (Let x be the independent variable and y be the dependent variable.)
figured it out
To find the equation of the line tangent to the graph of f(x) at x = 4, we first need to determine the slope of the tangent line.
The instantaneous rate of change (or slope) of f(x) at (4, -5) is given as 7. This means that when x = 4, the slope of the tangent line is 7.
Now, we can use the point-slope form of the equation of a line to find the equation of the tangent line.
The point-slope form is given by: y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope of the line.
Since the point on the tangent line is (4, -5) and the slope is 7, we can substitute these values into the point-slope form to obtain:
y - (-5) = 7(x - 4)
Simplifying the equation further:
y + 5 = 7x - 28
Subtracting 5 from both sides:
y = 7x - 33
Therefore, the equation of the line tangent to the graph of f(x) at x = 4 is y = 7x - 33.
To find the equation of the line tangent to the graph of f(x) at x = 4, we need to use the concept of derivatives.
The instantaneous rate of change of a function f(x) at a particular point is given by its derivative, which represents the slope of the tangent line at that point. So, we need to find the derivative of f(x) with respect to x and evaluate it at x = 4.
Let's denote the derivative of f(x) as f'(x) or dy/dx.
Given that the instantaneous rate of change of f(x) at (4, -5) is 7, we have:
f'(4) = 7
To find the equation of the tangent line, we need both the slope and a point on the line.
Using f'(4) = 7, we have found the slope of the tangent line at x = 4. Now we need to find the y-coordinate of the point of tangency.
Since the point (4, -5) lies on the graph of f(x), it also lies on the tangent line. Therefore, the y-coordinate of the point of tangency is -5.
Now we have the slope of the line (given by the derivative, which is 7) and a point on the line (x = 4, y = -5). We can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values:
y - (-5) = 7(x - 4)
Simplifying:
y + 5 = 7x - 28
y = 7x - 28 - 5
y = 7x - 33
Therefore, the equation of the line tangent to the graph of f(x) at x = 4 is y = 7x - 33.