Question

What is the equation of the graph below?

A graph shows a parabola that opens down and crosses the x axis near one and a half and four and a half.
A) y = − (x + 2)2 + 2
B) y = − (x − 3)2 + 2
C) y = (x − 2)2 + 2
D) y = (x + 3)2 + 2

Answer 1

B)
`y=-(x-3)^(2)+2`

Explanation:

We are given,

The graph of the parabola opens downwards.

This means that the leading co-efficient of the function will be negative.

So, options C and D are not possible.

Moreover, it is given that the graph cuts x-axis near
`1(1)/(2)` i.e.
`(3)/(2)` = 1.5 and
`4(1)/(2)` i.e.
`(9)/(2)` = 4.5.

i.e. At y=0, the value of x is near
`(3)/(2)` and
`(9)/(2)`.

A. So, in
`y=-(x+2)^(2)+2`, we put y =0.

i.e.
`(x+2)^(2)=2`

i.e.
`x=-2\pm √(2)`

i.e. x = 0.68 and x = 3.41

We see that this does not crosses the x-axis at the given points.

So, option A is wrong.

B. Also, in
`y=-(x-3)^(2)+2`, we put y =0.

i.e.
`(x-3)^(2)=2`

i.e.
`x=3\pm √(2)`

i.e. x = 1.68 and x = 4.41.

Thus, this equation cuts the x-axis near the given points i.e. 1.5 and 4.5.

Hence, the equation of the given graph is
`y=-(x-3)^(2)+2`.