You should always use the vertex and at least two points to graph each quadratic equation. A good choice for two points are the intercepts of the quadratic equation.
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The graph of a quadratic function is always a parabola. If you put the equation (or function) into vertex form, you can read off the coordinates of the vertex, and you know the shape and orientation (up/down) of the parabola.
The vertex form for a quadratic equation is y=a(x-h)^2+k.
look for the interceptions add these and divide it by 2 (that's the x vertex) for the yvertex you just have to fill in the x(vertex) however you can also use the formula -(b/2a)
A quadratic equation is an equation with the form: y=Ax2+Bx+C The most important point when graphing a parabola (the shape formed by a quadratic) is the vertex. The vertex is the maximum or minimum of the parabola. The x value of the vertex is equal to -B/(2A). Once you have the x value, just plug it back into the original equation to get the corresponding y value. The resulting ordered pair is the location of the vertex. A parabola will be concave up (pointed downward) if A is +. It will be concave down (pointed upward) if A is -. It is often helpful to find the zeroes of a function when graphing. This can be done by factoring or using the quadratic formula. For every n units away from the vertex on the x-axis, the corresponding y value goes up (or down) by n2*A. Parabolas are symetrical along the vertex, which means that if one point is n units from the vertex, the point -n units from the vertex has the same y value. As an example take the following quadratic: 2x2-8x+3 A=2, B=-8, and C=3 The x value of the vertex is -B/2A=-(-8)/(2*2)=2 By plugging 2 into the original equation we get that the vertex is at (2,-5) 3 units to the right (x=5) has a y value of -5+32*2=13. This means that 3 units to the left (x=-1) has the same y value (-1,13). If you need a clearer explanation, ask a math teacher.
In a quadratic equation, the vertex (which will be the maximum value of a negative quadratic and the minimum value of a positive quadratic) is in the exact center of any two x values whose corresponding y values are equal. So, you'd start by solving for x, given any y value in the function's range. Then, you'd solve for y where x equals the middle value of the two x's given in the previous. For example:y = x24 = x2x = 2, -2y = (0)2y = 0Which is, indeed, the vertex of y = x2