Question

What is the x-coordinate of the point on the graph of the equation y=2x+3 which is closest to the point (5,0)?

Answer 1

The closest distance means the distance square will be minimum as well.

so use
d^2 = (x - 5)^2 + ((2x+3) - 0)^2

Answer 2

Then try to minimize this.

Answer 3

So I used the distance formula r= sqrt((x1-x2)^2+(y1-y2)^2)
r=sqrt((x1-5)^2+(y1-0)^2) and followed up using y=2x+3
r=sqrt((x-5)^2+(2x+3)^2)

Answer 4

By the way, I'm using the calculator method, not calc.

Answer 5

You don't need calc or even a calculator to do this.

Answer 6

how else would you find the minimum?

Answer 7

You can't used -b/2a, can you?

Answer 8

The definition of a parabola is the distance between some point (focus) and some line (directrix), so we know we have a parabola. A parabola always has its critical point at the vertex.

Answer 9

Alright, then. How would you minimize this: r=sqrt((x-5)^2+(2x+3)^2)?

Answer 10

It leads to sqrt(5x^2+2x+34), by the way.

Answer 11

Sure, well first square both sides. We know that for all cases where r < r', then r^2 < r'^2

Answer 12

This is because we know r is positive.

Answer 13

So then we have:
r^2 = 5x^2+2x+34

Complete the square, and you get something of the form:
r^2 = a(x+h)^2 + k
We know the vertex is going to be (-h,k)

Answer 14

We'll know that -h is the x coordinate of the closest point, and put it into y=2x+3 to get the y coordinate

Answer 15

but can we ignore the ^2 on the distance when finding the vertex?

Answer 16

Okay, let me explain this again.

Suppose we know that r is the smallest possible distance.

Answer 17

This means that for all r', we know r < r'

Answer 18

Suppose we square both sides. Since we know that r and r' are non-negative (for the fact they are distances), we know that r^2 < r' * r < r'^2

Answer 19

In other worse, squaring will preserve order and the minimum square is still the minimum.

Answer 20

I just wasn't sure if the x-input value would be preserved.

Answer 21

Is completing the square too much? Are inequalities too much?

Answer 22

The x input doesn't need to be preserved.

Answer 23

The x input that minimizes r will minimize r^2

Answer 24

I understand the inequalities, though I'm a bit rusty on changing quadratic functions to vertex form.

Answer 25

That is, I don't know how to do it if the leading coefficient isn't 1. Help? :D

Answer 26

Divide the whole thing by 5.

Answer 27

now the leading coefficient is 1. Multiply the 5 back in when you're done.

Answer 28

Oh wait. Nevermind. Purplemath to the rescue!
Thanks for everything.

Answer 29

http://wio-math.herokuapp.com/notes/completing-the-square

Answer 30

Did you write that, or is it a frightful coincidence?

Answer 31

I wrote it a while ago

Answer 32

oh cool.

Answer 33

Like a couple months ago

Answer 34

Are you planning to finish the website?

Answer 35

I honestly do not know. Haven't worked on it in a while. My interest change a lot

Answer 36

This was incredibly strange though, given that completing the square was the only thing that was completed on the website and I just so happened to have broached this topic.

Answer 37

It had quadratic formula also, but I might have deleted it or the link to it.

Answer 38

A lot of time was just creating the infrastructure so that it would be easy to add in lessons. The infrastructure is there, but now it's hard to come up with concise lessons that don't feel awkward.

Answer 39

I blame mathjax.

Answer 40

It's not mathjax that is the problem.

Answer 41

I will take your word for it. Time to ask more questions!

Answer 42

I can pull up the code for that lesson, to show how simple it is.

Answer 43

```
extends ../note-layout

block vars
- var topic = 'Completing the Square'

block main
+section('Synopsis')
:markdown
**Completing the square** is the process of converting a quadratic expression
into its vertex form \\(a(x+h)^2+k\\) given its standard form \\(ax^2+bx+c\\).

One way is to evaluate equations of \\(h\\) and \\(k\\) given
the coefficients \\(a\\), \\(b\\), and \\(c\\).

They are \\(h=\frac b{2a}\\) and \\(k=c-\left(\frac b{2a}\right)^2\\).
.important
:mathjax
ax^2+bx+c = a\left(x+\frac b{2a}\right)^2+c-\left(\frac b{2a}\right)^2
:markdown
In the case where \\(a=1\\) and \\(c\\) is ignored,
we get \\(h=\frac b{2}\\) and \\(k=\left(\frac b{2}\right)^2\\).

These equations are a bit easier to remember and can still be used
even when \\(a\neq 1\\), so long as \\(a\\) is divided or factored out
before completing the square.
.important
:mathjax
x^2+bx = \left(x+\frac b{2}\right)^2-\left(\frac b{2}\right)^2
+section('Applications', {closed: true})
:markdown
* Finding the roots of a quadratic equation.
* Converting a quadratic equation vertex form.
* Converting an equation of an ellipse, hyperbola, or parabola into its standard equation.

```

Answer 44

What was the problem, then?