Standard form to vertex form

 

What is the square root of 9?

• You will say that it is 3, but actually it is 3 and (-3)

• 3x3=9

• (-3)x(-3)=9

• We is called the "plus/minus" and it is written as + 3.

 

We can find the x-intercepts from vertex form if we set y=0.

Let's try an example!

y=2(x-5)^2-50

• y=0

• 0=2(x-5)^2-50

• 0+50=2(x-5)^2-50+50

• 50=2(x-5)^2

• 50/2=(2(x-5)^2)/2

• 25=(x-5)^2

• Sqaure root both sides.

• + 5=x-5

• We need to find two x-intercepts. so first we do +5=x-5, and then we do -5=x-5.

• 5=x-5

• 5+5=x-5+5

• x=10

• Now we do -5=x-5 because we get + 5 when we square root 25, meaning +5 and -5.

• -5=x-5

• -5+5=x-5+5

• x=0

• We have the two x-intercepts of 0 and 10.

 

What if we are given standard form and we want to convert it into vertex form?

• We have to make it into a perfect square since vertex form is y=a(x-h)^2+k.

 

Let's go through one example!

y=x^2+8x+4

• Check if it can be factored.

• In this case it cannot be factored.

• Find the common factor of the first two terms which are x^2+8x.

• The common factor can only be a number, since it will become the "a" value, and the "a" value cannot have any variables.

• So there is no common factor in this case.

• Place brackets around the first two terms, so it becomes y=(x^2+8x)+4.

• Now change it into a perfect square.

• Remember that in perfect square the "b" value is equal to 2 multiplied by the square root of "a" and square root of "c".

• We are missing the "c" value, which has to be an integer.

• In order to find "c" we must divide 8(b) by 2 and square that number.

• 8/2 is 4, and square of 4 is 16.

• The "c" value is 16.

• You can check this by finding the square root of it, which is 4, and finding the square root of "a" which is 1, and we multiply these sqaure roots by 2.

• We get 2(1)(4)=8

• We end up with y=(x^2+8x+16)+4

• However, we cannot just change the equation like this; we need to balance this.

• If we have added 16, then we need to subtract it as well.

• We get y=(x^2+8x+16-16)+4

• The co-efficient outside the brackets (a) is 1, so we multiply it by 1, and we get 16.

• We bring the -16 outside the brackets and we get y=(x^2+8x+16x)-16+4

• We factor x^2+8x+16, which will end up as (x+4)^2.

• y=(x+4)^2-16+4.

• Collect like terms which are -16+4=-12

• Finally, we get y=(x+4)^2-12.

• We have converted standard form into vertex form, and we are ready to graph!

 

Important rule: For example, if we have y=3(x^2+8x+16)+4, then when we are balancing, we would not subtract 16, we would subtract 3(16)=48, since 16 is inside the bracket, and 3 is being multiplied with everything inside the bracket.

 

We would get y=3(x^2+8x+16-48)+4, and we would continue untill we convert it into vertex form.

 

Video time

 

 

 

 

 

 

 

 

 

 

 

 

Practice these...

1) y=2x^2-6x+1

2) y=2x^2+16x+3

3) y=-4x^2-8x-1

 

 

 

 

Answers are...

 

 

 

 

1) y=2(x-3)^2-17

2) y=2(x+4)^2-29

3) y=-4(x+1)^2+3