The derivative of the function is given as:
[tex]g^{\prime}(x)=\frac{x^2-16}{x-2}[/tex]
It is also given that g(3)=4.
Note that the Slope of a Tangent Line to a function at a point is the value of the derivative at that point.
Substitute x=3 into the derivative:
[tex]g^{\prime}(3)=\frac{3^2-16}{3-2}=\frac{9-16}{1}=\frac{-7}{1}=-7[/tex]
It follows that the slope of the tangent line at x=3 is -7.
Since it is given that g(3)=4, it implies that (3,4) is a point on the line.
Recall that the equation of a line with slope m, which passes through a point (x₁,y₁) is given by the point-slope formula as:
[tex]y-y_!=m(x-x_1)[/tex]
Substitute the point (x₁,y₁)=(3,4) and the slope m=-7 into the point-slope formula:
[tex]\begin{gathered} y-4=-7(x-3) \\ \Rightarrow y-4=-7x+21 \\ \Rightarrow y=-7x+21+4 \\ \Rightarrow y=-7x+25 \end{gathered}[/tex]
Hence, the equation of the tangent line to the graph of g at x=3 is y=-7x+25.
The required equation is y=-7x+25.
