Question

station 4. find the graph of f(x)=-x^{2}+2

Explanation:

Step1: Identify the form of the function

The function $f(x)=-x^{2}+2$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-1$, $b = 0$, and $c = 2$.

Step2: Determine the vertex

For a quadratic function $y=ax^{2}+bx + c$, the x - coordinate of the vertex is $x=-\frac{b}{2a}$. Substituting $a=-1$ and $b = 0$, we get $x = 0$. Substitute $x = 0$ into the function $f(x)$: $f(0)=-(0)^{2}+2=2$. So the vertex is $(0,2)$.

Step3: Determine the direction of the parabola

Since $a=-1<0$, the parabola opens downwards.

Step4: Find the x - intercepts

Set $y = f(x)=0$, so $-x^{2}+2 = 0$. Then $x^{2}=2$, and $x=\pm\sqrt{2}$. The x - intercepts are $x =-\sqrt{2}$ and $x=\sqrt{2}$.

Answer:

The graph is a parabola that opens downwards with vertex at $(0,2)$ and x - intercepts at $x =-\sqrt{2}$ and $x=\sqrt{2}$.