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

y= (blank)

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Acyclics

Answer:

y = (x+9)² - 4

Step-by-step 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)

View image Acyclics
chetankachana

Answer:

y = (x + 9)² - 4

Step-by-step explanation:

Hello!

The current form of the equation is in Standard Form: [tex]ax^2 + bx + c = 0[/tex]

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

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.

[tex](a + b)^2 = a^2 + 2ab + b^2\\(a - b)^2 = a^2 - 2ab + b^2[/tex]

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

  • [tex]y = x^2 + 18x + 77[/tex]

Replace y with 0

  • [tex]0 = x^2 + 18x + 77[/tex]

Complete the Square

  • [tex]0 + (\frac{18}{2})^2 = x^2 + 18x + 77 + (\frac{18}{2})^2\\0 + 81 = x^2 + 18x + 77 + 81[/tex]

Use the 81 to convert it into a Perfect Square. ([tex]a^2 + 2ab + b^2[/tex])

  • [tex]0 + 81 = x^2 + 18x + 81 + 77[/tex]
  • [tex]0 + 81 = (x + 9)^2 + 77[/tex]

Subtract 81 from both sides

  • [tex]0 = (x + 9)^2 - 4[/tex]
  • [tex]y = (x + 9)^2 - 4[/tex]

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

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 [tex](a + b)^2 = a^2 + 2ab + b^2[/tex], 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 [tex]y = a(x - h)^2 + k[/tex] the vertex is (h,k).

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

  • [tex]-\frac{b}{2a}[/tex]
  • [tex]-\frac{18}{2}[/tex]
  • [tex]-9[/tex]

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

  • [tex]y=x^2+18x+77[/tex]
  • [tex]y=(-9)^2+18(-9)+77[/tex]
  • [tex]y = 81 - 162 + 77[/tex]
  • [tex]y = -4[/tex]

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

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


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