Question

find the slope of the functions graph at the given point. then find an equation for the line tangent to the graph there.
f(x)=x² + 1, (-5,26)
what is the slope of the functions graph at the given point?
m = 27 (simplify your 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 \(y=-10x - 24\)