Question

find the slope of the functions graph at the given point. then find an equation for the line tangent to the graph there.
$f(x)=x^{2}+1,(-5,26)$
what is the slope of the functions graph at the given point?
$m=$ (simplify your 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 \(y = 27x + 109\).