axis of symmetry; Calculate the y-intercept; Use the axis of symmetry and the y-intercept to find an additional point on the graph; Graph the function
please help. This is college algebra not calculus. confuse
Given a quadratic function find: axis of symmetry, y-intercept. Use the axis of symmetry and the y-intercept to find an additional point on the graph then graph the function.
Let `y=ax^2+bx+c` e.g x^2+7x+12 (Here a=1,b=7,c=12)
(1) The axis of symmetry is the vertical line `x=-b/(2a)` . For the example `x=-7/2` .
(2) The y-intercept is where x=0, or c. For the example the y-intercept is 12.
(3) The line x=0 is 7/2 to the right of the axis of symmetry, so there is a point with y value of 12 when `x=-7/2-7/2=-7` . Thus there is a point (0,12) (the y-intercept) and a point (-7,12), a reflection of the intercept across the line of symmetry. (Imagine folding the paper along the line `x=-7/2` and pairing up points.)
(4) The graph is a "U" opening up (Since a>0). The example yields
** If the equation is given in another form there are other approaches. **