Question

graph $y = -2(x - 1)^2 - 3$.

Explanation:

Step1: Identify vertex form

The quadratic is in vertex form $y=a(x-h)^2+k$, where $(h,k)$ is the vertex. For $y=-2(x-1)^2-3$, $h=1$, $k=-3$, $a=-2$.
Vertex: $(1, -3)$

Step2: Determine parabola direction

Since $a=-2<0$, the parabola opens downward.

Step3: Find additional points

For $x=0$:

$y=-2(0-1)^2-3=-2(1)-3=-5$ → Point: $(0, -5)$

For $x=2$:

$y=-2(2-1)^2-3=-2(1)-3=-5$ → Point: $(2, -5)$

For $x=-1$:

$y=-2(-1-1)^2-3=-2(4)-3=-11$ (outside grid, skip)

For $x=3$:

$y=-2(3-1)^2-3=-2(4)-3=-11$ (outside grid, skip)

For $x=4$:

$y=-2(4-1)^2-3=-2(9)-3=-21$ (outside grid, skip)

Step4: Plot and connect points

Plot vertex $(1,-3)$, symmetric points $(0,-5)$ and $(2,-5)$, then draw a downward-opening parabola through these points.

Answer:

The parabola has vertex at $(1, -3)$, opens downward, and passes through points like $(0, -5)$ and $(2, -5)$. The graph is a downward-opening U-shape centered at $x=1$, with its lowest point at $(1, -3)$.