QuestionRewrite $y=x^{2}-6 x+5$ in vertex form. Is the vertex a max or min? Provide the coordinates. A. Max at $(-3,-4)$ B. Min at $(-3,-4)$ C. Min at $(3,-4)$ D. Max at $(3,-4)$

Studdy Solution

STEP 1

Assumptions1. The given equation is $y=x^{}-6 x+5$ . We need to rewrite it in vertex form which is $y=a(x-h)^{}+k$
3. The vertex of the parabola is at point $(h,k)$ 4. If $a>0$ , the parabola opens upwards and the vertex is a minimum. If $a<0$ , the parabola opens downwards and the vertex is a maximum.

STEP 2

The first step in completing the square is to rewrite the given equation in the form $y=a(x^{2}+bx)+c$ . Here, $a=1$ , $b=-6$ , and $c=5$ .
$y=x^{2}-6 x+5$

STEP 3

Next, we will complete the square. The term that completes the square is $(b/2)^{2}$ . Here, $b=-6$ , so $(b/2)^{2}=(-6/2)^{2}=9$ .

STEP 4

Add and subtract this term inside the brackets to complete the square.
$y=(x^{2}-6 x+9)-9+$

STEP 5

Now, simplify the equation. The term inside the brackets can be written as a square of a binomial.
$y=(x-3)^{2}-4$

STEP 6

Now, we have the equation in vertex form, $y=a(x-h)^{2}+k$ . Here, $a=1$ , $h=3$ , and $k=-4$ .

STEP 7

Since $a=1$ is greater than0, the parabola opens upwards. Therefore, the vertex is a minimum.

STEP 8

The coordinates of the vertex are $(h,k)=(3,-4)$ .
The solution is The vertex is a minimum at $(3,-4)$ .

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QuestionRewrite $y=x^{2}-6 x+5$ in vertex form. Is the vertex a max or min? Provide the coordinates. A. Max at $(-3,-4)$ B. Min at $(-3,-4)$ C. Min at $(3,-4)$ D. Max at $(3,-4)$

Studdy Solution

STEP 1

STEP 2

The first step in completing the square is to rewrite the given equation in the form $y=a(x^{2}+bx)+c$ . Here, $a=1$ , $b=-6$ , and $c=5$ .
$y=x^{2}-6 x+5$

STEP 3

Next, we will complete the square. The term that completes the square is $(b/2)^{2}$ . Here, $b=-6$ , so $(b/2)^{2}=(-6/2)^{2}=9$ .

STEP 4

Add and subtract this term inside the brackets to complete the square.
$y=(x^{2}-6 x+9)-9+$

STEP 5

Now, simplify the equation. The term inside the brackets can be written as a square of a binomial.
$y=(x-3)^{2}-4$

STEP 6

Now, we have the equation in vertex form, $y=a(x-h)^{2}+k$ . Here, $a=1$ , $h=3$ , and $k=-4$ .

STEP 7

Since $a=1$ is greater than0, the parabola opens upwards. Therefore, the vertex is a minimum.

STEP 8

The coordinates of the vertex are $(h,k)=(3,-4)$ .
The solution is The vertex is a minimum at $(3,-4)$ .

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QuestionRewrite y=x2−6x+5y=x^{2}-6 x+5y=x2−6x+5 in vertex form. Is the vertex a max or min? Provide the coordinates. A. Max at (−3,−4)(-3,-4)(−3,−4) B. Min at (−3,−4)(-3,-4)(−3,−4) C. Min at (3,−4)(3,-4)(3,−4) D. Max at (3,−4)(3,-4)(3,−4)

Studdy Solution

STEP 1

STEP 2

The first step in completing the square is to rewrite the given equation in the form y=a(x2+bx)+cy=a(x^{2}+bx)+cy=a(x2+bx)+c. Here, a=1a=1a=1, b=−6b=-6b=−6, and c=5c=5c=5.y=x2−6x+5y=x^{2}-6 x+5y=x2−6x+5

STEP 3

Next, we will complete the square. The term that completes the square is (b/2)2(b/2)^{2}(b/2)2. Here, b=−6b=-6b=−6, so (b/2)2=(−6/2)2=9(b/2)^{2}=(-6/2)^{2}=9(b/2)2=(−6/2)2=9.

STEP 4

Add and subtract this term inside the brackets to complete the square.y=(x2−6x+9)−9+y=(x^{2}-6 x+9)-9+y=(x2−6x+9)−9+

STEP 5

Now, simplify the equation. The term inside the brackets can be written as a square of a binomial.y=(x−3)2−4y=(x-3)^{2}-4y=(x−3)2−4

STEP 6

Now, we have the equation in vertex form, y=a(x−h)2+ky=a(x-h)^{2}+ky=a(x−h)2+k. Here, a=1a=1a=1, h=3h=3h=3, and k=−4k=-4k=−4.

STEP 7

Since a=1a=1a=1 is greater than0, the parabola opens upwards. Therefore, the vertex is a minimum.

STEP 8

The coordinates of the vertex are (h,k)=(3,−4)(h,k)=(3,-4)(h,k)=(3,−4).The solution is The vertex is a minimum at (3,−4)(3,-4)(3,−4).

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QuestionRewrite $y=x^{2}-6 x+5$ in vertex form. Is the vertex a max or min? Provide the coordinates. A. Max at $(-3,-4)$ B. Min at $(-3,-4)$ C. Min at $(3,-4)$ D. Max at $(3,-4)$

The first step in completing the square is to rewrite the given equation in the form $y=a(x^{2}+bx)+c$ . Here, $a=1$ , $b=-6$ , and $c=5$ .
$y=x^{2}-6 x+5$

Next, we will complete the square. The term that completes the square is $(b/2)^{2}$ . Here, $b=-6$ , so $(b/2)^{2}=(-6/2)^{2}=9$ .

Add and subtract this term inside the brackets to complete the square.
$y=(x^{2}-6 x+9)-9+$

Now, simplify the equation. The term inside the brackets can be written as a square of a binomial.
$y=(x-3)^{2}-4$

Now, we have the equation in vertex form, $y=a(x-h)^{2}+k$ . Here, $a=1$ , $h=3$ , and $k=-4$ .

Since $a=1$ is greater than0, the parabola opens upwards. Therefore, the vertex is a minimum.

The coordinates of the vertex are $(h,k)=(3,-4)$ .
The solution is The vertex is a minimum at $(3,-4)$ .