For a function [tex]\( h \)[/tex], we are given that [tex]\( h(5) = 2 \)[/tex] and
[tex]\[ h^{\prime}(5) = -6 \][/tex]

What's the equation of the tangent line to the graph of [tex]\( h \)[/tex] at [tex]\( x = 5 \)[/tex]?



Answer :

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]