Question

Put the equation y=x^2+18x+77 into the form y=(x-h)^2+k:

y= (blank)

Explain your answer!

Show your work!

No spam answers!

No incorrect answers!

No nonsense answers!

Thanks!

Answer 1

y = (x+9)² - 4

Explanation:

Important concepts that must be known to solve this problem:

Quadratic Form : y = ax² + bx + c

X value of vertex = -b/2a

To find y value of vertex plug in value of x into equation

Vertex Form : y = (x-h)² + k where (h,k) is at vertex

Objective : To convert the equation to vertex form from quadratic form

Here, we are given the equation y = x² + 18x + 77

And we are wanting to convert this equation to vertex form or y = (x-h)² + k

To do so we must find the vertex

We know that the x value of the vertex = -b/2a

Remember quadratic form is y = ax² + bx + c and we have x² + 18x + 77

No value takes the place of a and 18 takes the place of b

So we can say a = 1 and b = 18

X value = -b/2a = -18/2 = -9

Now to find the y value of the vertex we plug in the value of x

y = x² + 18x + 77

==> plug in x = -9

y = (-9)² + 18(-9) + 77

==> evaluate exponent

y = 81 + 18(-9) + 77

==> multiply 18 by -9

y = 81 - 162 + 77

==> combine terms

y = -4

So we have x = -9 and y = -4

Meaning the vertex is at (-9,-4)

Now we can plug in the values into vertex form

Recall that vertex form is y = (x-h)² + k

We have (h,k) = (-9,-4) so h = -9 and k = -4

Plugging in these values of h and k

We acquire y = (x - (-9))² + (-4)

Simplifying this we get y = (x+9)² - 4

We can check our answer by graphing

If you look at the attached image you can see that both equations result in the same graph (they overlap)

Answer 2

y = (x + 9)² - 4

Explanation:

Hello!

The current form of the equation is in Standard Form:
`ax^2 + bx + c = 0`

We want to convert it into Vertex Form:
`y = a(x - h)^2 + k`

We can do this by Completing The Square.

What is Completing The Square?

Completing the square is when you solve a quadratic by having a Perfect Square on one side. Essentially you want to factor a perfect Square Trinomial by adding your own value to both sides so it makes it possible. You can only complete the square when the coefficient of the x-variable is 1.

`(a + b)^2 = a^2 + 2ab + b^2\\(a - b)^2 = a^2 - 2ab + b^2`

To create a perfect square from Standard Form (ax² + bx + c = 0):

• Take the b-value
• Divide it by 2
• Square it
• Add it to both Sides

Now, let's try it with our equation.

Complete the Square

• 
  `y = x^2 + 18x + 77`

Replace y with 0

• 
  `0 = x^2 + 18x + 77`

Complete the Square

• 
  `0 + ((18)/(2))^2 = x^2 + 18x + 77 + ((18)/(2))^2\\0 + 81 = x^2 + 18x + 77 + 81`

Use the 81 to convert it into a Perfect Square. (
`a^2 + 2ab + b^2`)

• 
  `0 + 81 = x^2 + 18x + 81 + 77`
• 
  `0 + 81 = (x + 9)^2 + 77`

Subtract 81 from both sides

• 
  `0 = (x + 9)^2 - 4`
• 
  `y = (x + 9)^2 - 4`

Therefore, the equation in Vertex Form is
`y = (x + 9)^2 - 4`.

Difference between Perfect Square Factoring and Regular Factoring

Factoring this equation regularly will give us (x + 7)(x + 11) = y.

But we want both factors to be the same, so that it is the square of one factor.

Since there is another term being added that will not give us a perfect square, we have to add a completely new term to factor it into a Perfect Square (81), and factor using that term rather than 77. This is how we got it:

y = (x + 9)² - 4

Expanding (x + 9)² will give us x² + 18x + 81, not the same as x² + 18x + 77. That's why we subtracted 4 at the end.

Another simple way to Complete the Square:

Ignore the term with no variable completely (77) and focus on x² + 18x.

If you want to factor it into a perfect square, we have to add something to it.

Using the Formula
`(a + b)^2 = a^2 + 2ab + b^2`, we know that "A" is 1, as the coefficient of x is 1. "2" times "1" times "b" is 18, so b would be 9.

Add the square of 9 to both sides to get 81. Then, factor it using the formula.

• y + 81 = x² + 18x + 81
• y + 81 = (x + 9)²

Subtract the 81 from the left-hand side, and finally, add the 77 that we left out at the start.

• y = (x + 9)² - 4

It's pretty much the same concept as the first method, but once we organize it a bit better, it works faster and is not as chaotic.

There is also a third way: Finding the Vertex

Vertex Form is named what it is for a reason.

It's because the Vertex is inside the Equation. In
`y = a(x - h)^2 + k` the vertex is (h,k).

We can find the x-value of the vertex from the Standard Form: ax² + bx + c

• 
  `-(b)/(2a)`
• 
  `-(18)/(2)`
• 
  `-9`

Plug in the value of -9 into the equation to find the y-value of the vertex

• 
  `y=x^2+18x+77`
• 
  `y=(-9)^2+18(-9)+77`
• 
  `y = 81 - 162 + 77`
• 
  `y = -4`

So the vertex would be (-9,-4).

Plugging that into the formula, we get
`y = (x +9)^2 - 4`. A is 1 since the coefficient of x is 1.