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.\nf(x)=x² + 1, (-5,26)\nwhat is the slope of the functions graph at the given point?\nm = 27 (simplify your answer.)
Answer
Explanation:
Step1: Find the derivative of the function
The function is (f(x)=x^{2}+1). Using the power rule ((x^n)^\prime = nx^{n - 1}), the derivative (f^\prime(x)=2x).
Step2: Calculate the slope at the given point
The (x) - coordinate of the point ((-5,26)) is (x=-5). Substitute (x = - 5) into (f^\prime(x)): (m=f^\prime(-5)=2\times(-5)=-10).
Step3: Use the point - slope form to find the tangent line equation
The point - slope form is (y - y_1=m(x - x_1)), where ((x_1,y_1)=(-5,26)) and (m=-10). Substitute the values: (y - 26=-10(x + 5)). Expand: (y-26=-10x-50). Add 26 to both sides: (y=-10x-24).
Answer:
The slope (m = - 10) and the equation of the tangent line is (y=-10x - 24)