Graph the following state and show the a h and k values.
y=2(x+3)^2-4

This equation is in vertex form. Vertex form is y=a(x-h)^2+k, where (h,k) is the vertex. In this equation, h= -3, and k= -4. So, the vertex would be at (-3,-4). A tells us if the parabola will be positive or negative. In this case, it is positive, so the parabola opens upward. Then, you can solve for x-intercept by letting y=0 and solving for x. Then, you can find the y-intercept by setting x=0 and solving for y. Finally, graph the parabola.

paul's with a y-intercept of $ 2856

Aquadratic function is one of the form f(x)=ax2 + bx + c l, where a,b, and c are numbers with are not equal to zero.