I've done this question a few times but I can't seem to figure out what I'm doing wrong…
In the question, there is a statement and then a graph.
The question states:
Applying the Mean Value Theorem with $a = 2$, $b = 7$, and $c = 4$. What is
the equation of the tangent line at 4?
Then the figure below the question shows part of circle with the points, $(2, 3)$, $(4, 6)$, $(7, 7)$ and a tangent line passing the point, $(4, 6)$, which, from what I understand, is point $c$.
The answer is a fill in the blanks,
$y =$ ______
So, first I wrote out the equation for The Mean Value Theorem:
$${f'(c) = \frac{f(b) – f(a)}{b – a}}$$
Then I plugged in the values:
$${f'(4) = \frac{f(7) – f(2)}{7 – 2}}$$
Using the points on the graph, I plug in the f(x) values
$${f'(4) = \frac{7 – 3}{5}}$$
$${f'(4) = \frac{4}{5}}$$
$\frac{4}{5}$, however, is not the answer because I am not looking for $y'$, I'm looking for $y$.
What do I do next? How do I find y?
Best Answer
Using the point parallel of the tangent line, point c, and the slope, plug them into the equation $y = mx + b$
$${y = 4/5x + b}$$ $$ 6 = 4/5(4) + b$$ $$14/5 = b$$ $$y = 4/5x + 14/5$$
That is the complete and correct equation for $y$.