Question 1196808
Sketch the following functions by first solving for the x intercepts using the quadratic formula.  If there are no x intercepts then use the method explained in class...this makes use of the fact that raising or lowering a parabola by some amount does not change the x coordinate of the vertex.  So...remove the constant from the right side, factor what's left to determine x intercepts.  The average of those is the x coordinate of the vertex.  Then substitute that back into the original equation to solve for the y coordinate of the vertex. 
y = 3x^2+15x+30   AND   y = -x^2+4x-8
<pre>First, as stated by one of the people who responded, {{{y=3(x+25/4)^2-75/4}}}-----------you can read the vertex and solve for the roots"
<font size = 4><font color = red><b>IS NOT</font></font></b> the vertex form of the equation of: {{{matrix(1,3, 3x^2 + 15x + 30, "=", 0)}}}

{{{matrix(1,3, y, "=", 3x^2 + 15x + 30)}}}
Although prompted to use the quadratic equation formula to find the x intercepts, it's not necessary. Calculating the discriminant, {{{matrix(1,7, b^2 - 4ac, or, 15^2 - 4(3)(30), "=", 225 - 360, "=", - 135)}}}, it's absolutely clear that this equation's solutions are IMAGINARY, and so, DOES NOT
have any x-intercepts.

{{{matrix(1,3, y, "=", - x^2 + 4x - 8)}}}
Although prompted to use the quadratic equation formula to find the x intercepts, it's not necesary. Calculating the discriminant, {{{matrix(1,7, b^2 - 4ac, or, 4^2 - 4(- 1)(- 8), "=", 16 - 32, "=", - 16)}}}, it's absolutely clear that this equation's solutions are IMAGINARY, and so, DOES NOT
have any x-intercepts.

I have NO IDEA what method was explained in YOUR class!!

                                          {{{matrix(1,3, y, "=", 3x^2 + 15x + 30)}}}

Removing the constant, the above becomes: {{{matrix(2,3, y, "=", 3x^2 + 15x + 30 - 30, y, "=", 3x^2 + 15x)}}}

                                          {{{matrix(2,3, 0, "=", 3x^2 + 15x, 0, "=", 3x(x + 5))}}}
                                           0 = 3x     or      0 = x + 5
                                           0 = x      or    - 5 = x 
                               x-intercepts: 0 and - 5, or (0, 0) and (- 5, 0)
x-coordinate of vertex of: {{{matrix(1,8, 3x^2 + 15x + 30, ":", (0 + - 5)/2, "=", - 5/2, ",", "or", highlight(- 2.5))}}} 
y-coordinate of vertex of: {{{matrix(1,9, 3x^2 + 15x + 30, "=", 3(- 2.5)^2 + 15(- 2.5) + 30, "=", 3(6.25) - 37.5 + 30, "=", 18.75 - 37.5 + 30, "=", highlight(11.25))}}}
Coordinates of vertex of: 3x<sup>2</sup> + 15x + 30: (- 2.5, 11.25). 
      
                                          {{{matrix(1,3, y, "=",  - x^2 + 4x - 8)}}}

Removing the constant, the above becomes: {{{matrix(2,3, y, "=", - x^2 + 4x - 8 + 8, y, "=", - x^2 + 4x)}}}

                                          {{{matrix(2,3, 0, "=", - x^2 + 4x, 0, "=", - x(x - 4))}}}
                                           0 = - x     or      0 = x - 4
                                           0 = x       or      4 = x 
                               x-intercepts: 0 and 4, or (0, 0) and (4, 0)
x-coordinate of vertex of: {{{matrix(1,7, - x^2 + 4x - 8, "=", (0 + 4)/2, "=", 4/2, "=", highlight(2))}}} 
y-coordinate of vertex of: {{{matrix(1,7, - x^2 + 4x - 8, "=", - (2)^2 + 4(2) - 8, "=", - 4 + 8 - 8, "=", highlight(- 4))}}}
Coordinates of vertex of: - x<sup>2</sup> + 4x - 8: (2, - 4).</pre>