The given function is
[tex]f(x)=15-x^2^{}[/tex]
First, we have to find f(x+h), which consists of adding h to the x-variable.
[tex]f(x+h)=15-(x+h)^2[/tex]
Then, we use the functions we have in the slope definition.
[tex]\lim _{h\to0}(\frac{f(x+h)-f(x)}{h})=\lim _{h\to0}(\frac{15-(x+h)^2-(15-x^2)}{h})[/tex]
Solve the notable product and multiply by the negative sign in front.
[tex]\lim _{h\to0}(\frac{15-(x^2+2xh+h^2)-15+x^2}{h})=\lim _{h\to0}(\frac{15-x^2-2xh-h^2-15+x^2}{h})[/tex]
Add like terms.
[tex]\lim _{h\to0}(\frac{-2xh-h^2}{h})[/tex]
Factor out the greatest common factor h.
[tex]\lim _{h\to0}\frac{h(-2x-h)}{h}=\lim _{h\to0}(-2x-h)[/tex]
At last, evaluate the expression when h = 0.
[tex]\lim _{h\to0}(-2x-h)=-2x-0=-2x[/tex]
The slope is given by the equation -2x.
Let's evaluate the expression when x = 2 to find the value of the slope.
[tex]m=-2x=-2(2)=-4[/tex]
(a) Therefore, the slope is -4.
To find the equation for the tangent, we have to use the point-slope formula.
[tex]\begin{gathered} y-y_1=m(x-x_1);\text{ where} \\ x_1=2,y_1=11,m=-4 \end{gathered}[/tex]
Use the values of the coordinates and slope to find the equation.
[tex]\begin{gathered} y-11=-4(x-2) \\ y-11=-4x+8 \\ y=-4x+8+11 \\ y=-4x+19 \end{gathered}[/tex]
(b) Therefore, the equation for the tangent to the curve is y = -4x+19.
(c)
The image below shows the function f(x) and the tangent.
