Question

What is the equation of the graph below?

Answer 1

Answer:

B) $y=-(x-3)^{2}+2$

Step-by-step 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 $1\frac{1}{2}$ i.e. $\frac{3}{2}$ = 1.5 and $4\frac{1}{2}$ i.e. $\frac{9}{2}$ = 4.5.

i.e. At y=0, the value of x is near $\frac{3}{2}$ and $\frac{9}{2}$.

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

i.e. $(x+2)^{2}=2$

i.e. $x=-2\pm \sqrt{2}$

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. $x=3\pm \sqrt{2}$

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 $y=-(x-3)^{2}+2$.