To find the equation of the tangent line to the graph of a function [tex]\( h \)[/tex] at a specific point, you can use the point-slope form of the equation of a line. The point-slope form is given by:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\( m \)[/tex] is the slope of the line.
Given the values:
- [tex]\( h(5) = 2 \)[/tex], which means the point of tangency is [tex]\((5, 2)\)[/tex],
- [tex]\( h'(5) = -6 \)[/tex], which indicates the slope of the tangent line at [tex]\( x = 5 \)[/tex].
With [tex]\((x_1, y_1) = (5, 2)\)[/tex] and [tex]\( m = -6 \)[/tex], we can substitute these values into the point-slope form equation as follows:
[tex]\[
y - 2 = -6(x - 5)
\][/tex]
Now, let's rearrange this equation to get it into the slope-intercept form [tex]\( y = mx + b \)[/tex]:
1. Start with the point-slope form:
[tex]\[
y - 2 = -6(x - 5)
\][/tex]
2. Distribute the slope [tex]\( -6 \)[/tex] on the right-hand side:
[tex]\[
y - 2 = -6x + 30
\][/tex]
3. Add 2 to both sides of the equation to isolate [tex]\( y \)[/tex]:
[tex]\[
y = -6x + 30 + 2
\][/tex]
4. Simplify the right-hand side:
[tex]\[
y = -6x + 32
\][/tex]
Therefore, the equation of the tangent line to the graph of [tex]\( h \)[/tex] at [tex]\( x = 5 \)[/tex] is:
[tex]\[
y = -6x + 32
\][/tex]
