Math / Algebra

Question10 pts
The vertex form of the quadratic function $f(x)=x^{2}-6 x+5$ is $f(x)=a(x-h)^{2}+k$ . What is the value of $a$ ? $\square$ 1
What is the value of $h$ ? $\square$ $-5$
What is the value of $k$ ? $\square$ $-24$

Studdy Solution

STEP 1

What is this asking? We need to rewrite the quadratic function $f(x) = x^2 - 6x + 5$ in vertex form, $f(x) = a(x - h)^2 + k$ , and find the values of $a$ , $h$ , and $k$ . Watch out! Don't forget to factor out the coefficient of the $x^2$ term if it's not 1!
Also, be super careful with signs when completing the square.

STEP 2

1. Prepare for completing the square
2. Complete the square
3. Identify a, h, and k

STEP 3

Let's group the terms with $x$ together: $f(x) = (x^2 - 6x) + 5$ This gets us ready to complete the square!

STEP 4

To complete the square, we take half of the coefficient of the $x$ term, which is $\frac{-6}{2} = \mathbf{-3}$ , and square it: $(-3)^2 = \mathbf{9}$ .
This is our magic number!

STEP 5

We add and subtract our magic number inside the parentheses to keep the equation balanced: $f(x) = (x^2 - 6x + 9 - 9) + 5$

STEP 6

Now, we can rewrite the first three terms as a perfect square trinomial: $f(x) = ((x - 3)^2 - 9) + 5$ Remember, $(x-3)^2 = x^2 - 6x + 9$ !

STEP 7

Finally, let's simplify: $f(x) = (x - 3)^2 - 9 + 5$ $f(x) = (x - 3)^2 - 4$

STEP 8

Now, we have our quadratic in vertex form: $f(x) = 1 \cdot (x - 3)^2 - 4$ .
Comparing this to $f(x) = a(x - h)^2 + k$ , we can easily see that $a = \mathbf{1}$ , $h = \mathbf{3}$ , and $k = \mathbf{-4}$ .

STEP 9

$a = 1$ , $h = 3$ , and $k = -4$ .

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Question10 pts
The vertex form of the quadratic function $f(x)=x^{2}-6 x+5$ is $f(x)=a(x-h)^{2}+k$ . What is the value of $a$ ? $\square$ 1
What is the value of $h$ ? $\square$ $-5$
What is the value of $k$ ? $\square$ $-24$

Studdy Solution

STEP 1

STEP 2

1. Prepare for completing the square
2. Complete the square
3. Identify a, h, and k

STEP 3

Let's group the terms with $x$ together: $f(x) = (x^2 - 6x) + 5$ This gets us ready to complete the square!

STEP 4

To complete the square, we take half of the coefficient of the $x$ term, which is $\frac{-6}{2} = \mathbf{-3}$ , and square it: $(-3)^2 = \mathbf{9}$ .
This is our magic number!

STEP 5

We add and subtract our magic number inside the parentheses to keep the equation balanced: $f(x) = (x^2 - 6x + 9 - 9) + 5$

STEP 6

Now, we can rewrite the first three terms as a perfect square trinomial: $f(x) = ((x - 3)^2 - 9) + 5$ Remember, $(x-3)^2 = x^2 - 6x + 9$ !

STEP 7

Finally, let's simplify: $f(x) = (x - 3)^2 - 9 + 5$ $f(x) = (x - 3)^2 - 4$

STEP 8

Now, we have our quadratic in vertex form: $f(x) = 1 \cdot (x - 3)^2 - 4$ .
Comparing this to $f(x) = a(x - h)^2 + k$ , we can easily see that $a = \mathbf{1}$ , $h = \mathbf{3}$ , and $k = \mathbf{-4}$ .

STEP 9

$a = 1$ , $h = 3$ , and $k = -4$ .

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Studdy Solution

STEP 1

STEP 2

1. Prepare for completing the square2. Complete the square3. Identify *a*, *h*, and *k*

STEP 3

Let's **group** the terms with xxx together: f(x)=(x2−6x)+5f(x) = (x^2 - 6x) + 5f(x)=(x2−6x)+5 This gets us ready to **complete the square**!

STEP 4

To complete the square, we take half of the coefficient of the xxx term, which is −62=−3\frac{-6}{2} = \mathbf{-3}2−6​=−3, and square it: (−3)2=9(-3)^2 = \mathbf{9}(−3)2=9.This is our **magic number**!

STEP 5

We **add** and **subtract** our magic number inside the parentheses to keep the equation balanced: f(x)=(x2−6x+9−9)+5f(x) = (x^2 - 6x + 9 - 9) + 5f(x)=(x2−6x+9−9)+5

STEP 6

Now, we can rewrite the first three terms as a perfect square trinomial: f(x)=((x−3)2−9)+5f(x) = ((x - 3)^2 - 9) + 5f(x)=((x−3)2−9)+5 Remember, (x−3)2=x2−6x+9(x-3)^2 = x^2 - 6x + 9(x−3)2=x2−6x+9!

STEP 7

Finally, let's simplify: f(x)=(x−3)2−9+5f(x) = (x - 3)^2 - 9 + 5f(x)=(x−3)2−9+5 f(x)=(x−3)2−4f(x) = (x - 3)^2 - 4f(x)=(x−3)2−4

STEP 8

Now, we have our quadratic in vertex form: f(x)=1⋅(x−3)2−4f(x) = 1 \cdot (x - 3)^2 - 4f(x)=1⋅(x−3)2−4.Comparing this to f(x)=a(x−h)2+kf(x) = a(x - h)^2 + kf(x)=a(x−h)2+k, we can easily see that a=1a = \mathbf{1}a=1, h=3h = \mathbf{3}h=3, and k=−4k = \mathbf{-4}k=−4.

STEP 9

a=1a = 1a=1, h=3h = 3h=3, and k=−4k = -4k=−4.

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1. Prepare for completing the square
2. Complete the square
3. Identify a, h, and k

Let's group the terms with $x$ together: $f(x) = (x^2 - 6x) + 5$ This gets us ready to complete the square!

To complete the square, we take half of the coefficient of the $x$ term, which is $\frac{-6}{2} = \mathbf{-3}$ , and square it: $(-3)^2 = \mathbf{9}$ .
This is our magic number!

We add and subtract our magic number inside the parentheses to keep the equation balanced: $f(x) = (x^2 - 6x + 9 - 9) + 5$

Now, we can rewrite the first three terms as a perfect square trinomial: $f(x) = ((x - 3)^2 - 9) + 5$ Remember, $(x-3)^2 = x^2 - 6x + 9$ !

Finally, let's simplify: $f(x) = (x - 3)^2 - 9 + 5$ $f(x) = (x - 3)^2 - 4$

Now, we have our quadratic in vertex form: $f(x) = 1 \cdot (x - 3)^2 - 4$ .
Comparing this to $f(x) = a(x - h)^2 + k$ , we can easily see that $a = \mathbf{1}$ , $h = \mathbf{3}$ , and $k = \mathbf{-4}$ .

$a = 1$ , $h = 3$ , and $k = -4$ .