Question

I tried to answer it but it says its wrong

Answer 1

We have

`f(x)=(1)/(2)(x-1)^2-2`

This is a parabola in the vertex form, we can see that the vertex of the parabola is

`(1,-2)`

Because of the vertex form:

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

There the vertex is

`(h,k)`

Then, we can already plot one point of the parabola

`(1,-2)`

Let's also find the y-intercept, doing x = 0

`\begin{gathered} f(x)=(1)/(2)(x-1)^(2)-2 \\ \\ f(0)=(1)/(2)(0-1)^2-2 \\ \\ f(0)=(1)/(2)(-1)^2-2 \\ \\ f(0)=(1)/(2)-2 \\ \\ f(0)=-(3)/(2) \end{gathered}`

Therefore we have another point

`(0,-3/2)`

We can also find the zeros of that parabola doing y = 0

`\begin{gathered} 0=(1)/(2)(x-1)^2-2 \\ \\ (1)/(2)(x-1)^2=2 \\ \\ (x-1)^2=4 \\ \\ (x-1)^2=4=\begin{cases}x-1={2} \\ x-1=-2\end{cases} \\ \\ \end{gathered}`

Solving the two linear equations we get the zeros

The zeros are 3 and -1. We have all the points to draw a perfect parabola!

Draw the parabola