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 derivative of $f(x)=x^{2}+1$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $f^\prime(x)=2x$.

Step2: Evaluate the derivative at the given x - value

We want to find the slope at the point $(5,26)$. Substitute $x = 5$ into $f^\prime(x)$. So $m=f^\prime(5)=2\times5 = 10$.

Step3: Find the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(5,26)$ and $m = 10$.
Substitute these values: $y-26=10(x - 5)$.
Expand: $y-26=10x-50$.
Solve for $y$: $y=10x - 24$.

Answer:

The slope $m = 10$. The equation of the tangent line is $y=10x - 24$.