Math / Algebra

Question3/10
Write $y=x^{2}-18 x+52$ in vertex form. $y=(x-9)^{2}+52$ $y=(x-9)^{2}-29$ $y=(x-9)^{2}+113$ $y=(x+9)^{2}-29$

Studdy Solution

STEP 1

What is this asking? Rewrite the equation $y = x^2 - 18x + 52$ from its standard form into its vertex form, which helps us easily see the vertex of the parabola. Watch out! Don't mess up those signs when completing the square!
A little slip-up can lead to a completely different parabola.

STEP 2

1. Complete the Square
2. Write in Vertex Form

STEP 3

Let's look at the $x$ terms in our equation, $x^2 - 18x$ .
To complete the square, we need to find a value that makes this part of the equation a perfect square trinomial.

STEP 4

Remember, we take half of the coefficient of our $x$ term, which is $-18$ , and square it!
Half of $-18$ is $-9$ , and $(-9)^2$ is 81.
This is our magic number!

STEP 5

We add our magic number, $81$ , to the $x$ terms, but to keep the equation balanced, we also subtract it.
This is like adding zero in a clever way! $y = x^2 - 18x + \mathbf{81} - \mathbf{81} + 52$

STEP 6

Now, the first three terms form a perfect square trinomial, which we can factor: $y = (x - 9)^2 - 81 + 52$

STEP 7

Combine those constant terms, $-81 + 52 = \mathbf{-29}$ : $y = (x - 9)^2 - 29$

STEP 8

Our equation is now in vertex form, which is generally written as $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.

STEP 9

In our case, $a = 1$ , $h = 9$ , and $k = -29$ .
So, the vertex of our parabola is $(9, -29)$ .
Awesome!

STEP 10

The vertex form of the given equation is $y = (x - 9)^2 - 29$ .

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Question3/10
Write $y=x^{2}-18 x+52$ in vertex form. $y=(x-9)^{2}+52$ $y=(x-9)^{2}-29$ $y=(x-9)^{2}+113$ $y=(x+9)^{2}-29$

Studdy Solution

STEP 1

What is this asking? Rewrite the equation $y = x^2 - 18x + 52$ from its standard form into its vertex form, which helps us easily see the vertex of the parabola. Watch out! Don't mess up those signs when completing the square!
A little slip-up can lead to a completely different parabola.

STEP 2

1. Complete the Square
2. Write in Vertex Form

STEP 3

Let's look at the $x$ terms in our equation, $x^2 - 18x$ .
To complete the square, we need to find a value that makes this part of the equation a perfect square trinomial.

STEP 4

Remember, we take half of the coefficient of our $x$ term, which is $-18$ , and square it!
Half of $-18$ is $-9$ , and $(-9)^2$ is 81.
This is our magic number!

STEP 5

We add our magic number, $81$ , to the $x$ terms, but to keep the equation balanced, we also subtract it.
This is like adding zero in a clever way! $y = x^2 - 18x + \mathbf{81} - \mathbf{81} + 52$

STEP 6

Now, the first three terms form a perfect square trinomial, which we can factor: $y = (x - 9)^2 - 81 + 52$

STEP 7

Combine those constant terms, $-81 + 52 = \mathbf{-29}$ : $y = (x - 9)^2 - 29$

STEP 8

Our equation is now in vertex form, which is generally written as $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.

STEP 9

In our case, $a = 1$ , $h = 9$ , and $k = -29$ .
So, the vertex of our parabola is $(9, -29)$ .
Awesome!

STEP 10

The vertex form of the given equation is $y = (x - 9)^2 - 29$ .

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Question3/10Write y=x2−18x+52y=x^{2}-18 x+52y=x2−18x+52 in vertex form. y=(x−9)2+52y=(x-9)^{2}+52y=(x−9)2+52 y=(x−9)2−29y=(x-9)^{2}-29y=(x−9)2−29 y=(x−9)2+113y=(x-9)^{2}+113y=(x−9)2+113 y=(x+9)2−29y=(x+9)^{2}-29y=(x+9)2−29

Studdy Solution

STEP 1

STEP 2

1. Complete the Square2. Write in Vertex Form

STEP 3

Let's look at the xxx terms in our equation, x2−18xx^2 - 18xx2−18x.To complete the square, we need to find a value that makes this part of the equation a perfect square trinomial.

STEP 4

Remember, we take half of the coefficient of our xxx term, which is −18-18−18, and square it!Half of −18-18−18 is −9-9−9, and (−9)2(-9)^2(−9)2 is **81**.This is our **magic number**!

STEP 5

We add our **magic number**, 818181, to the xxx terms, but to keep the equation balanced, we *also* subtract it.This is like adding zero in a clever way! y=x2−18x+81−81+52 y = x^2 - 18x + \mathbf{81} - \mathbf{81} + 52 y=x2−18x+81−81+52

STEP 6

Now, the first three terms form a perfect square trinomial, which we can factor: y=(x−9)2−81+52 y = (x - 9)^2 - 81 + 52 y=(x−9)2−81+52

STEP 7

Combine those constant terms, −81+52=−29-81 + 52 = \mathbf{-29}−81+52=−29: y=(x−9)2−29 y = (x - 9)^2 - 29 y=(x−9)2−29

STEP 8

Our equation is now in **vertex form**, which is generally written as y=a(x−h)2+ky = a(x - h)^2 + ky=a(x−h)2+k, where (h,k)(h, k)(h,k) is the **vertex** of the parabola.

STEP 9

In our case, a=1a = 1a=1, h=9h = 9h=9, and k=−29k = -29k=−29.So, the **vertex** of our parabola is (9,−29)(9, -29)(9,−29).Awesome!

STEP 10

The vertex form of the given equation is y=(x−9)2−29y = (x - 9)^2 - 29y=(x−9)2−29.

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Question3/10
Write $y=x^{2}-18 x+52$ in vertex form. $y=(x-9)^{2}+52$ $y=(x-9)^{2}-29$ $y=(x-9)^{2}+113$ $y=(x+9)^{2}-29$

1. Complete the Square
2. Write in Vertex Form

Let's look at the $x$ terms in our equation, $x^2 - 18x$ .
To complete the square, we need to find a value that makes this part of the equation a perfect square trinomial.

Remember, we take half of the coefficient of our $x$ term, which is $-18$ , and square it!
Half of $-18$ is $-9$ , and $(-9)^2$ is 81.
This is our magic number!

We add our magic number, $81$ , to the $x$ terms, but to keep the equation balanced, we also subtract it.
This is like adding zero in a clever way! $y = x^2 - 18x + \mathbf{81} - \mathbf{81} + 52$

Now, the first three terms form a perfect square trinomial, which we can factor: $y = (x - 9)^2 - 81 + 52$

Combine those constant terms, $-81 + 52 = \mathbf{-29}$ : $y = (x - 9)^2 - 29$

Our equation is now in vertex form, which is generally written as $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.

In our case, $a = 1$ , $h = 9$ , and $k = -29$ .
So, the vertex of our parabola is $(9, -29)$ .
Awesome!

The vertex form of the given equation is $y = (x - 9)^2 - 29$ .