As Jason C points out in the comments, the version you posted is most likely the best choice, due to its superior precision near the edge cases:
float lerp(float a, float b, float f)
{
return a * (1.0 - f) + (b * f);
}
If we disregard from precision for a while, we can simplify the expression as follows:
a(1 − f) × (b − a)
= a − af + bf
= a + f(b − a)
Which means we could write it like this:
float lerp(float a, float b, float f)
{
return a + f * (b - a);
}
In this version we’ve gotten rid of one multiplication, but lost some precision.