Question

graph $f(t) = -3(t - 2)^2 + 1$.
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Explanation:

Step1: Identify vertex form parameters

The quadratic is in vertex form $f(t)=a(t-h)^2+k$, where $a=-3$, $h=2$, $k=1$.

Step2: Locate the vertex

Vertex is at $(h,k)=(2, 1)$.

Step3: Determine parabola direction

Since $a=-3<0$, parabola opens downward.

Step4: Find additional points

Calculate $f(1)=-3(1-2)^2+1=-3+1=-2$, so point $(1,-2)$.
Calculate $f(3)=-3(3-2)^2+1=-3+1=-2$, so point $(3,-2)$.
Calculate $f(0)=-3(0-2)^2+1=-12+1=-11$, so point $(0,-11)$.
Calculate $f(4)=-3(4-2)^2+1=-12+1=-11$, so point $(4,-11)$.

Answer:

The graph is a downward-opening parabola with vertex at $(2, 1)$, passing through points $(1,-2)$, $(3,-2)$, $(0,-11)$, $(4,-11)$, and symmetric about the line $t=2$. Plot these points and draw a smooth parabolic curve through them.