Question

I tried to answer it but it says its wrong

Answer 1

We have

$f(x)=\frac{1}{2}(x-1)^2-2$

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

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

Therefore we have another point

$(0,-3/2)$

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

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

Solving the two linear equations we get the zeros

$\begin{gathered} x-1=2\Rightarrow x=3 \\ x-1=-2\operatorname{\Rightarrow}x=-1 \end{gathered}$

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

Draw the parabola