Ways to Draw Graph of the Quadratic Function
Graph of the function `y=ax^2+bx+c`, where `a!=0` is parabola. To draw it three methods are used, that will be illustrated in following example.
Example . Draw graph of the function `y=-0.5x^2-x+4`.
First way: finding of coordinates `(x_0,y_0)` of vertex of parabola using following formulas: `x_0=-b/(2a)`; `y_0=(4ac-b^2)/(4a)`.
Here `a=-0.5,b=-1,c=4`. Therefore, `x_0=-(-1)/(2*(-0.5))=-1`, `y_0=(4*(-0.5)*4-1)/(4*(-0.5))=9/2`. So, `(-1;9/2)` is vertex of parabola. To graph, let's find a couple more points on parabola: for example, `(0;4),(1;5/2),(2;0)`. Draw vertex of parabola, obtained points and points, that are symmetric to them about axis of parabola. Now, connect all points - we obtained graph of the parabola (see figure).
Second way: drawing parabola using points whose y-coordinate equals free member of quadratic function.
Let's find points of graph, whose y-coordinate equals free member of quadratic cfunction. For this we need to solve equation `-0.5x^2-x+4=4`, i.e. `0.5x^2+x=0`. This equations has two roots: `x_1=0,x_2=-2`.
So, we found two points of graph: A(0;4) and B(-2;4).
Since points A and B lie on graph and have same y-coordinate then they are symmetric about axis of symmetry of parabola, thus axis passes through middle of segment AB perpendicular to it. x-coordinate of point A equals 0, x-coordinate of point B equals -2, therefore, x-coordinate of middle is `(0+(-2))/2=-1`, thus x=-1 is axis of symmetry of parabola. Now, `y(-1)=-0.5*(-1)^2-(-1)+4=9/2`. Therefore, point `C(-1,9/2)` is vertex of parabola (i.e. the only point of parabola that lies on its axis of symmetry). Finally, we draw points A,B,C; connect them with line and obtain graph of the parabola (see figure).
Third way: drawing parabola using roots of quadratic function.
Equation `-0.5x^2-x+4=0` has two roots: `x_1=-4` and `x_2=2`. Therefore, we know two points of parabola: D(-4;0) and E(2;0). Since axis of symmetry is perpendicular to segment DE and passes through middle of it, then x-coordinate of middle is `(-4+2)/2=-1`. So, x=-1 is line of symmetry. Therefore, vertex of parabola is point `C(-1;9/2)`. Using three points D,E and C we draw graph (see figure).