To find the derivative of the function [tex]\( h \)[/tex] at the point [tex]\((-2, -4)\)[/tex], denoted as [tex]\( h'(-2) \)[/tex], we need to determine the slope of the tangent line at that point. The slope of the tangent line can be calculated using the coordinates of the given points through which the line passes.
The coordinates given are:
- Point 1: [tex]\((-2, -4)\)[/tex]
- Point 2: [tex]\((1, 5)\)[/tex]
The slope [tex]\( m \)[/tex] of the line passing through these two points is calculated using the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Substituting the coordinates into the formula:
[tex]\[
m = \frac{5 - (-4)}{1 - (-2)}
\][/tex]
Simplify the numerator and the denominator:
[tex]\[
m = \frac{5 + 4}{1 + 2}
\][/tex]
[tex]\[
m = \frac{9}{3}
\][/tex]
[tex]\[
m = 3
\][/tex]
Therefore, the slope of the tangent line at the point [tex]\((-2, -4)\)[/tex], which is also the value of the derivative [tex]\( h'(-2) \)[/tex], is:
[tex]\[
h'(-2) = 3.0
\][/tex]
So,
[tex]\[
h'(-2) = 3.0
\][/tex]
