Math, asked by otfyoungmiller, 1 year ago

Complete the square to rewrite y = x2 – 6x + 15 in vertex form. Then state whether the vertex is a maximum or minimum and give its coordinates. A. Maximum at (–3, 6) B. Minimum at (–3, 6) C. Maximum at (3, 6) D. Minimum at (3, 6)

Answers

Answered by isyllus
16

Answer:

Option C is correct. Maximum at (3,6)

Step-by-step explanation:

We are given a equation of parabola in standard form, y=x^2-6x+15

we need to convert into vertex form, y=a(x-h)^2+k

Using completing square method,

y=x^2-6x+15                      

( Add and subtract square of half of coefficient of x)

y=x^2-6x+9-9+15

y=(x-3)^2+6                    a^2-2ab+b^2=(a-b)^2

This would be vertex form of parabola,

Vertex: (3,6)

The leading coefficient of parabola is positive.

Therefore, At vertex we will get maximum.

Maximum at (3,6)

Thus, Option C is correct. Maximum at (3,6)

Answered by parmesanchilliwack
28

Answer:

C. Maximum at (3, 6)

Step-by-step explanation:

Here, the given equation is,

y=x^2-6x+15 -------(1)

Which is the equation of parabola,

Since, the vertex form of parabola is,

y=a(x-h)^2+k

Where, (h,k) is the vertex of the parabola,

If a > 0 then the value of y is minimum at (h,k),

While, if a < 0, then the value of y is maximum at (h,k),

Now, from equation (1),

y=x^2-6x+15

y=x^2-6x+15 +9-9

y=(x-3)^2+15-9

y=(x-3)^2+6

By comparing this equation with the vertex form of the parabola,

We get, (h,k) = (3,6)

And, a = 1 > 0

Hence,  y is minimum at (3,6).

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