If f(2)=3 and f'(2)=5, write an equation for the tangent line to the graph of y=f(x) at the point where x=2.
You need to write the equation of the tangent line, hence you need at least two points to write this equation suct that:
`y - y_1 = (y_2 - y_1)/(x_2 - x_1) * (x - x_1)`
Notice that the problem provides the derivative of the function at the point x = 2. You need to link this information to the fact that the derivative of a function at a point is the tangent line to the graph of function at the point, such that:
f'(x) = m (m denotes the slope of the line or the tangent of angle made by line to x axis)
You need to remember what tangent function is:
tan `alpha` = opp side/adj side
You need to find the length of orthogonal projection to y axis and x axis such that:
`l_1 ` = opp side`= y_2 - y_1`
`` `l_2` = adj side `= x_2 - x_1`
Hence, `m = (y_2 - y_1)/(x_2 - x_1)`
Notice that the point (2,3) is on the graph of f(x) and on the tangent line such that:
`m = (y_2 - y_1)/(x_2 - x_1) = 5`
You may write the new form of equation of tangent line:
`y = 5(x-2) + 3 =gt y = 5x - 7`
Hence, the equation of tangent line is y = 5x - 7.
Remember point-slope form for a line:
We can fill in all we need from the given information.
y - 3 = 5(x - 2)
This answer is fine, but if you need to you can put it into slope-intercept form:
y-3 = 5x - 10
y = 5x - 7