Verified answer

your teacher has probably already told you, but here goes.

The graph of a parabola is always going to be a U shape. If you think of it as a cup, if the x^2 coefficient is positive, it "holds water" and curves up, if negative, then it is inverted and curves down. To find the x-intercepts, you set y=0 and solve for the x roots (if there are no real roots, then the parabola is completely above or below the x-axis, if there is one real root, then the parabola is tangent to the x-axis). To find the y-intercept, set x=0 and solve for y. That will give you some main points on the curve. If you need a few more, plug in integer values from x=-3 to x=3, and you will be able to fill out your curve. This problem can also be simplified:

y = (2x)(x+2) which gives you the x roots of 0 and -2, without having to use the quadratic formula. This means when y is 0, the curve passes through the points (-2,0) and (0,0). Since the coefficient "a" is positive (as in y= ax^2 + bx + c) then it curves up, and would have a low point halfway between -2 and 0, or at the point x=-1. Plug in -1 for x to get the value of y at that minimum point.

Hope this helps, without doing the whole thing for you.

One way to graph a parabola is to just find the vertex point, then find a pair or two of points on either side to roughly draw the rest of it. When you have a parabola equation in the form of y-k = a(x-h)², then (h,k) is the vertext point. So use completing the square to get the equation into that form:

y = 2x²+4x

y = 2(x²+2x)

y + 2 = 2(x²+2x+1)

y + 2 = 2(x+1)²

So the vertex point is at (-1, -2). When y=0, you get x = 0 or -2, THis means (0,0) and (-2,0) are also points on the parabola. When you plot these, it's obvious where the rest of the parabola roughly is.

y=2(x^2+2x) = 2(x+1)^2 -2 So the vertice is (-1,-2) The axis is x=-1

The x intercepts are x=0 and x= -2 and the y intercept is (0,0)

If you need more points for graphing give values to x and calculate y