1. You have a function f with the equation y = f(x)
2. You have a point P(a, f(a)) which is placed on the graph of f.
3. Now the x-value changes a little bit. The new x-value is (a+h)
4. You get another point on the graph Q(a+h, (f(a+h))
5. Now you can calculate the slope between these 2 points:
6. This is the difference quotient for x = a.
7. Because you can use this equation for every x of the domain of f normally the equation is written like this:
. Difference quotient
According to your problem x = 0 therefore m = -h
. First derivation.
According to your problem x = 0 therefore f'(0) = 0
If x = -2 the f(-2) = 5 that means the point P(-2, 5) lies on the graph of f.
The gradient of the graph at x = -2 must be the same as the slope of the tangent in P.
Now use point-slope-formula of a straight line to get the equation of the tangent:
Plug in all known values into this equation and solve for y:
A horizontal line is a parallel to the x-axis that means it has the slope zero:
The slope of the tangent is the same as the gradient of the function:
Now f'(x) = 0. Therefore:
Plug in these x-values into the equation of the function to get the second coordinate of the tangent points.