Question

Let f(X) be the square of the distance from the oint (2,1) to a point (x, 3x+2) on the line y=3x+2. Show that f(x) is a quadratic function, and find its minimum value by completing the square.

Answer 1

f(x) = square of distance between points (2, 1) and (x, 3x+2)

Answer 2

use the distance formula and write down f(x)

Answer 3

is that it?

Answer 4

wat did u get after applying distance formula ?

Answer 5

I got root (-10x^2 +13)

Answer 6

ok, lets see if its correct :)

Answer 7

we have the distance formula ,
distance between two points = \(\large \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)

Answer 8

distance between points (2, 1) and (x, 3x+2) =

\(\sqrt{(x-2)^2+(3x+2-1)^2}\)

\(\sqrt{(x-2)^2+(3x+1)^2}\)

\(\sqrt{x^2+4-4x+ 9x^2+1+6x}\)

\(\sqrt{10x^2+2x+5}\)

Answer 9

f(x) = square of distance between points (2, 1) and (x, 3x+2)

Answer 10

so,

f(x) = \((\sqrt{10x^2+2x+5})^2\)

= \(10x^2+2x+5\)

Answer 11

thats our required function f(x),
clearly, its a quadratic (why ?)

Answer 12

Oh I see, I foiled it wrong thank you so much!

Answer 13

np :) you still need to find its minimum value...

Answer 14

Its minimum value would be the lowest point on the graph c:

Answer 15

Yes, u knw how to find it ha ? :)

Answer 16

Can you review over how to for me?

Answer 17

ok sure :)
we simply need to change the quadratic to form : \(\large a(x-h)^2+\color{red}{k}\)

\(\color{red}{k}\) is our required minimum value

Answer 18

\(10x^2+2x+5\)

factor out 10

\(10(x^2+2/10x+5/10)\)

Answer 19

oh okay thank you! I know now, 49/10

Answer 20

\(10x^2+2x+5\)

factor out 10

\(10(x^2+2/10x+5/10)\)

\(10(x^2+2(1/10)x+ 1/100 - 1/100 + 5/10)\)

\(10((x+1/10)^2 - 1/100 + 5/10)\)

\(10((x+1/10)^2 - 1/100 + 50/100)\)

\(10((x+1/10)^2 + 49/100)\)

\(10(x+1/10)^2 + 10 \times 49/100\)

\(10(x+1/10)^2 + 49/10\)

Answer 21

Yes, \(\color{red}{k}\) = 49/10, so thats our min value !