The slope of the tangent to the graph of a function at x=x₀ can be estimated by finding the slope of a line that passes through two points near to x₀.
For instance, evaluate the function f at x₀+h and x₀. The slope will be given by:
[tex]m=\frac{f(x_0+h)-f(x_0)}{(x_0+h)-(x_0)}=\frac{f(x_0+h)-f(x_0)}{h}[/tex]For small values of h, this slope will approach the slope of the tangent line.
For the given function:
[tex]f(x)=2x^3+x^2+23[/tex]Use x₀=2 and h=0.01. Then, we must evaluate f at x=2.01 and at x=2:
[tex]\begin{gathered} f(2)=2(2)^3+(2)^2+23=43 \\ f(2.01)=2(2.01)^3+(2.01)^2+23=43.281302 \end{gathered}[/tex]Then, the slope of the tangent line is approximately:
[tex]m\approx\frac{f(2.01)-f(2)}{0.01}=\frac{43.281302-43}{0.1}=28.1302[/tex]We can get better approximations by choosing smaller values for h. For instance, if h=0.001, we would get:
[tex]m\approx\frac{f(2.001)-f(2)}{0.001}=\frac{43.028013-43}{0.001}=28.013\ldots[/tex]Therefore, the slope of the line tangent to the graph of y=2x³+x²+23 is approximately 28.