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
`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.
Since g'(5) = 4, then, the slope of the line that touches the graph of g(x) is 4. So,
Then, apply the point-slope form to get the equation of the tangent line.
So, plug-in the slope of the tangent line (m=4), and its point of tangency (5,-3).
And, isolate the y.
Therefore, the equation of the tangent line is `y=4x - 23` .