The tangent line will always be a straight line, we can use slope intercept form to create that line.

y = mx + b.

We know that the derivative at x = 5 equals 4, f'(5) = 4. So the slope at x = 5 is 4. y = 4x...

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The tangent line will always be a straight line, we can use slope intercept form to create that line.

y = mx + b.

We know that the derivative at x = 5 equals 4, f'(5) = 4. So the slope at x = 5 is 4.

y = 4x + b

Now we just have to solve for b, and then we will be finished. We plug in our point (5, -3) for x and y, and then we solve for b.

-3 = 4*5 + b

b = -23

So the equation for the line tangent to our function at the point (5, -3) is

y = 4x -23

The notation

`g(5) =-3 `

can be express as an order pair. Its equivalent order pair is (5,-3). So the tangent line touches the graph of g(x) at (5,-3).

Moreover, derivative of a function is the slope of the tangent line.

`m=g'(x)`

Since g'(5) = 4, then, the slope of the line that touches the graph of g(x) is 4. So,

m=4

Then, apply the point-slope form to get the equation of the tangent line.

`y-y_1=m(x-x_1)`

So, plug-in the slope of the tangent line (m=4), and its point of tangency (5,-3).

`y-(-3)=4(x-5)`

And, isolate the y.

`y+3=4x-20`

`y=4x-23`

**Therefore, the equation of the tangent line is `y=4x - 23` .**