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Identify all the points that are on the following quadratic function: y = x2 - 3x (0,3) (0,0) (2,-2) (-1,4) (-3,15)

2 Answers

4 votes
B,C,D
Three answers. 2nd,3rd,4th
User Shiyan Xu
by
8.0k points
6 votes

Answer:

The points on the quadratic function
y=x^(2)-3x are :


(0,0),
(2,-2) and
(-1,4)

Explanation:

We have the following quadratic function :


y=x^(2)-3x

If we want to see if a particular point is on the quadratic function we need to replace each pair
(x,y) in the quadratic function expression and check the equality.

For example :

We have the following point :
(x,y)=(1,2)

If we replace by
x=1 and
y=2 in the quadratic function :


y=x^(2)-3x


2=1^(2)-3(1)


2=1-3


2=-2

The final expression is false, therefore the point
(1,2) is not on the quadratic function.

Now let's work with the points we were given.

The first point is
(0,3) ⇒ If we replace in the quadratic function :


y=x^(2)-3x


3=0^(2)-3(0)


3=0

We conclude that the point
(0,3) is not in the quadratic function.

The second point is
(0,0)


y=x^(2)-3x


0=0^(2)-3(0)


0=0

This point is on the quadratic function.

The third point is
(2,-2)

If we replace in the quadratic function :


y=x^(2)-3x


-2=(2)^(2)-3(2)


-2=4-6


-2=-2

The point
(2,-2) is on the quadratic function.

The fourth point is
(-1,4)


y=x^(2)-3x


4=(-1)^(2)-3(-1)


4=1+3


4=4

Therefore the point
(-1,4) is on the quadratic function.

Finally, we have the point
(-3,15)

If we replace in the quadratic function :


y=x^(2)-3x


15=(-3)^(2)-3(-3)


15=9+9


15=18

This point is not on the quadratic function.

We conclude that the points
(0,0),(2,-2) and
(-1,4) are on the quadratic function
y=x^(2)-3x

User IngisKahn
by
7.2k points

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