### Home > CALC > Chapter 4 > Lesson 4.1.3 > Problem4-38

4-38.

On your paper, sketch a graph of $f(x) = x^3 + 3x^2-45x + 8$.

1. Find the slope of the line tangent to the curve at $x = −2$.

Recall that the slope of the tangent line at $x = a$ is also known as $f^\prime(a)$.

2. Find the point on the curve where the slope is the smallest (steepest negative slope). What is the name of this point?

You are looking for the location where $f^\prime(x)$ is at its lowest (or at a minimum). To find this location, you will need to find the location ($x$-value) of the lowest point (the vertex) on the graph of $f^\prime(x)$. Note: $f^\prime(x)$ is a parabola, so an Algebra I student could complete this part of the task.

Use the $x$-value of the vertex of $f^\prime(x)$, to evaluate the coordinate point on $f(x)$ where the slope is the steepest.