consider the function g(x)=-3(x+2)^2-4. Graph h(x)=g(x-1).

Answer 1

h(x) = g(x-1) = -3(x-1+2)² -4 Just replace x by x-1

h(x) = -3(x+1)² -4

The vertex or maximum of the parabola is (-1, 4) and it opens downward

Let x=0 and solve for the y-intercept y=h(0) = -3(1)-4 = -7 that's the point (0,-7)

Let h=0 and solve for the x-intercept(s). there should be two of them

0=-3(x+1)² -4

(x+1)² = 4/3 take the square roots of both sides

x+1 =+ or - 2/sqr3

x= -1 + or - 2/sqr3

=-1 + or - 2sqr3/3

x= -1-2sqr3/3 -1+2sqr3/3

= about -2.155, 0.155

the x-intercepts are the points (-2.155,0) and (0.155,0)

That's 4 points, just draw a smooth parabolic curve through them, symmetric about the vertical line x=-1. The parabola to the right of x=-1 is a mirror image of the parabola to the left of x=-1