Find k such that the line y = 5x + 3 is tangent to the graph of the function below. y=ksqrt(x)?
what does k=?
- Seamus OLv 710 years agoFavorite Answer
y = k√x
y' = k / 2√x
Now if the given line is tangent to y = k√x
then the slope of the line equals k / 2√x
so k / 2√x = 5
k = 10√x [eqn 1]
Now to find the point where tangent touches the curve:
k√x = 5x + 3 [eqn 2]
Subs [eqn 1] into [eqn 2] → 10√x . √x = 5x + 3
10x = 5x + 3
5x = 3
x = 3/5
Now subs x = 3/5 into [eqn 1] → k = 10√(3/5)