Question

Let f(x) be the square of the distance from the point (2,1) to a point (x,3x+2) on the line y=3x+2. Find the minimum value of f(x) using either algebraic techniques or graphing techniques.

Answer 1

Did you find the distance square between the points that were given?

Answer 2

sqrt((2-x)^2+(1-3x+2)^2)

Answer 3

well if you wanted to write that way the only bad thing I see is you didn't write 1-3x-2 in that one ( )

Answer 4

in order words you would have
\[d=\sqrt{(2-x)^2+(1-[3x+2])^2}\]
you need to distribute that -

Answer 5

\[d=\sqrt{(2-x)^2+(1-3x-2)^2}\]

Answer 6

Anyways distance squared would look like this:
\[d^2=(2-x)^2+(1-3x-2)^2\]
And it told us to rename this f(x)
so
\[f(x)=(2-x)^2+(1-3x-2)^2\]

Answer 7

Now you should realize the degree here is 2 so
you should be able to write it in y=ax^2+bx+c
or y=a(x-h)^2+k <---which will give the min of the parabola

Answer 8

You need simplify 1-3x-2 first

Answer 9

then do some multiply and combining of like terms