Question

graph $y = 4(x - 1)^2 - 3$.

Explanation:

Step1: Identify vertex form

The given equation is in vertex form $y=a(x-h)^2+k$, where $a=4$, $h=1$, $k=-3$.

Step2: Find vertex point

The vertex is $(h,k) = (1, -3)$.

Step3: Find axis of symmetry

Axis of symmetry is $x=h$, so $x=1$.

Step4: Calculate additional points

For $x=0$: $y=4(0-1)^2-3=4(1)-3=1$, point $(0,1)$
For $x=2$: $y=4(2-1)^2-3=4(1)-3=1$, point $(2,1)$
For $x=-1$: $y=4(-1-1)^2-3=4(4)-3=13$, point $(-1,13)$
For $x=3$: $y=4(3-1)^2-3=4(4)-3=13$, point $(3,13)$

Step5: Plot and connect points

Plot the vertex, symmetric points, then draw a parabola opening upward (since $a>0$).

Answer:

The graph is an upward-opening parabola with vertex at $(1, -3)$, axis of symmetry $x=1$, passing through points $(0,1)$, $(2,1)$, $(-1,13)$, $(3,13)$ (and other symmetric points as needed to form the curve).