find the slope of the functions graph at the given point. then find an equation for the line tangent to the…

find the slope of the functions graph at the given point. then find an equation for the line tangent to the graph there.\n$f(x)=x^{2}+1,(-5,26)$\nwhat is the slope of the functions graph at the given point?\n$m=$ (simplify your answer.)

find the slope of the functions graph at the given point. then find an equation for the line tangent to the graph there.\n$f(x)=x^{2}+1,(-5,26)$\nwhat is the slope of the functions graph at the given point?\n$m=$ (simplify your answer.)

Answer

Explanation:

Step1: Find the derivative of the function

The function is (f(x)=x^{3}+1). Using the power rule ((x^{n})^\prime = nx^{n - 1}), the derivative (f^\prime(x)=3x^{2}).

Step2: Evaluate the derivative at the given (x) - value

The (x) - value of the point ((- 3,28)) is (x=-3). Substitute (x = - 3) into (f^\prime(x)): (f^\prime(-3)=3\times(-3)^{2}). Since ((-3)^{2}=9), then (f^\prime(-3)=3\times9 = 27). So the slope (m = 27).

Step3: Use the point - slope form of a line

The point - slope form of a line is (y - y_{1}=m(x - x_{1})), where ((x_{1},y_{1})=(-3,28)) and (m = 27). Substitute these values into the formula: (y - 28=27(x+3)). Expand the right - hand side: (y - 28=27x + 81). Add 28 to both sides to get the equation of the tangent line: (y=27x+109).

Answer:

The slope (m = 27) and the equation of the tangent line is (y = 27x + 109).