If the instantaneous rate of change of f(x) at (2,-4) is 5, write the equation of the line tangent to the graph of f(x) at x=2.
slope = 5 at x = 2 so
y = 5 x + b
-4 = 5 (2) + b
-4 = 10 + b
b = -14
so
y = 5 x -14
you have a point and a slope, so just use the point-slope form, which you learned in Algebra I:
y+4 = 5(x-2)
To write the equation of the line tangent to the graph of f(x) at x=2, we need to find the slope of the tangent line.
The instantaneous rate of change of f(x) at (2,-4) is given as 5, which represents the slope of the tangent line at that point.
So, the slope of the tangent line is 5.
Now, we can use the point-slope form of a line to write the equation of the tangent line. The point-slope form is given as:
y - y1 = m(x - x1)
Here, (x1, y1) represents the point (2, -4), and m represents the slope of the tangent line, which is 5.
Plugging the values into the equation, we get:
y - (-4) = 5(x - 2)
Simplifying, we have:
y + 4 = 5x - 10
Finally, rearranging the equation to slope-intercept form, we get:
y = 5x - 14
Therefore, the equation of the line tangent to the graph of f(x) at x=2 is y = 5x - 14.
To write the equation of the line tangent to the graph of f(x) at x=2, we need two pieces of information: the slope of the tangent line and the point it passes through.
The instantaneous rate of change of f(x) at (2, -4) is given as 5. This means that the slope of the tangent line at x=2 is 5.
Now, let's find the point that the line passes through. We are given that the point is (2, -4), so we already have the x-coordinate. We need to find the corresponding y-coordinate.
To do that, we'll use the equation of the graph of f(x), denoted as y = f(x). Since the point (2, -4) lies on the graph, we can substitute the x-coordinate (2) into the equation to find the y-coordinate:
y = f(2)
So, our equation becomes:
-4 = f(2)
Now, we have the point (2, -4) that the line passes through and the slope (5) of the tangent line. We can use the point-slope form of the equation of a line to write the equation:
y - y₁ = m(x - x₁)
where (x₁, y₁) is the point on the line and m is the slope.
Substituting the values we have:
y - (-4) = 5(x - 2)
Simplifying:
y + 4 = 5x - 10
Now, let's isolate y:
y = 5x - 10 - 4
y = 5x - 14
The equation of the line tangent to the graph of f(x) at x=2 is y = 5x - 14.