202k views
3 votes
Please help me if you know how to do this

Please help me if you know how to do this-example-1
Please help me if you know how to do this-example-1
Please help me if you know how to do this-example-2
User Nafeesa
by
8.3k points

1 Answer

2 votes

Question 17

Answer: choice C) y = (x-2)^2 - 3

--------------------

Step-by-step explanation:

We know that a = 1 so it matches with the 'a' value in y = x^2. Otherwise, the shapes wouldn't match up and they wouldn't be congruent.

The vertex we want is (h,k) = (2,-3)

Therefore,

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

y = 1(x-2)^2 + (-3)

y = (x-2)^2 - 3

================================================

Question 18

Answer: choice B) min point; -4

--------------------

Step-by-step explanation:

The equation y = 3(x-1)^2 - 4 is in the form y = a(x-h)^2 + k which is known as vertex form.

a = 3 being positive tells us the graph opens upward and we have a min point. There is no max point as it goes on forever upward.

The min point is k = -4 which is the y coordinate of the vertex.

================================================

Question 19

Answer: Choice B) y = (1/2)x^2 + 3x - 5

--------------------

Step-by-step explanation:

The equation opens up, so the value of 'a' must be positive. This narrows our choices to A or B.

We eliminate choice A because that graph is more narrow compared to y = x^2. Choice B is wider and flatter compared to y = x^2 due to 1/2 being smaller than 1.

User Kalinka
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories