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Question
graph the function $f(x) = -3x^2$. plot the vertex. then plot another point on the parabola. if you make a mistake, you can erase your parabola by selecting the second point and placing it on top of the first.
Step1: Find the vertex
For $f(x)=ax^2+bx+c$, vertex at $x=-\frac{b}{2a}$. Here $a=-3, b=0$, so $x=0$. Substitute $x=0$: $f(0)=-3(0)^2=0$. Vertex is $(0,0)$.
Step2: Find a second point
Choose $x=1$. Substitute into $f(x)$: $f(1)=-3(1)^2=-3$. Point is $(1,-3)$.
Step3: Use symmetry for third point
Parabola is symmetric over $x=0$, so $x=-1$ gives $f(-1)=-3(-1)^2=-3$. Point is $(-1,-3)$.
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- Plot the vertex at $(0, 0)$.
- Plot the point $(1, -3)$ (and $(-1, -3)$ for symmetry), then draw a downward-opening parabola passing through these points.