lcx=- 3.5-sqrt(10.25) & (I) x=- 3.5+sqrt(10.25) & (II)
(I), (II): Subtract term
lx=-6.70156... x=-0.29843...
(I), (II): Round to 2 decimal place(s)
lx_1≈-6.70 x_2≈ -0.30
The roots are -6.70 and - 0.30. To sketch the parabola we need a few more points. Examining the function, we notice that it has a constant of 2, which means it intercepts the y-axis at y=2. Another useful point is the function's vertex. By rewriting the function in graphing form, we can find its coordinates.
Graphing Form:& y=(x-a)^2+ b
Vertex:& (a, b)
If we rewrite the function to match this form exactly, we can identify the coordinates of its vertex. Notice that when we completed the square we were able to do one part of this procedure.
Now we can identify the coordinates of the vertex.
Function:& y=(x-(-3.5))^2+( -10.25)
Vertex:& (-3.5, -10.25)
The vertex coordinates are (-3.5,-10.25). With this information, we can sketch the parabola.
