Question

graph the quadratic equation on the provided grid. 11. $f(x)=(x - 0)^2+3$

Explanation:

Step1: Identify the vertex form

The quadratic function $f(x)=(x - 0)^2+3$ is in vertex - form $y=a(x - h)^2+k$, where $(h,k)$ is the vertex. Here $h = 0$ and $k = 3$, so the vertex is $(0,3)$.

Step2: Find the y - intercept

Set $x = 0$ in $f(x)$. Since $f(0)=(0 - 0)^2+3=3$, the y - intercept is $(0,3)$ (which is the same as the vertex in this case).

Step3: Find some additional points

Let $x = 1$, then $f(1)=(1 - 0)^2+3=1 + 3=4$, so the point is $(1,4)$.
Let $x=-1$, then $f(-1)=(-1 - 0)^2+3=1 + 3=4$, so the point is $(-1,4)$.
Let $x = 2$, then $f(2)=(2 - 0)^2+3=4 + 3=7$, so the point is $(2,7)$.
Let $x=-2$, then $f(-2)=(-2 - 0)^2+3=4 + 3=7$, so the point is $(-2,7)$.

Step4: Plot the points

Plot the vertex $(0,3)$ and the additional points $(\pm1,4),(\pm2,7)$ on the grid and draw a smooth parabola passing through these points. The parabola opens upward because $a = 1>0$.

Answer:

Plot the vertex at $(0,3)$ and points $(\pm1,4),(\pm2,7)$ and draw an upward - opening parabola.