The graph of f’ is shown above. the points (2,6)(2,6) and (4,18)(4,18)are on the graph of f. which of the following is an equation of the line tangent to the graph of f at x=2?

For a given function f(x), the slope of the tangent line to a particular point (x₁, y₁) on f(x) is given by the first derivative of f(x), called f'(x), evaluated in the x-value of that point, so the slope is: f'(x₁).

Here the answer is: y = 5*x - 4

We know:

Remember that a general line is written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

Then to find the slope, we need to look at the graph of f' and find f'(2).

We can see that f'(2) = 5

Then the line is something like:

y = 5*x + b

Now to find the value of b, we can use the fact that this line is tangent to the point (2, 6), this means that when x = 2, we must have y = 6.

We can replace these in the line equation to get:

6 = 5*2 + b

6 = 10 + b

6 - 10  = b = -4

Then the equation of the line is:

y = 5*x - 4

If you want to learn more, you can read:

link

y = 5x-4

Explanation:

The point (2,5) is on the f ' (x) graph. This means x = 2 and f ' (x) = 5.

The slope of the tangent line is m = 5 when x = 2.

We are told that the point (2,6) is on the f(x) curve. So x = 2 and y = 6

Use m = 5, x = 2, y = 6 to find the value of b

y = mx+b

6 = 5*2+b

6 = 10+b

10+b = 6

b = 6-10

b = -4

The equation of the tangent line at x = 2 is therefore y = 5x-4

idk

step-by-step explanation: