1.  Determine the maximum value of the quadratic relation: y = -3x 2 + 12x + 29  We can’t factor this, so what do we do?  Look for the axis of symmetry, since the vertex lies on it ◦ We just need 2 points with the same y-value, and find the x-value in between ◦ Let’s try y = 29

2.  29 = -3x 2 + 12x + 29  0 = -3x 2 + 12x  0 = -3x(x – 4) so either x = 0 or x = 4  The axis of symmetry must be at x = 2  So, the x-coordinate of the vertex is x = 2  To find the y-coordinate, plug in x = 2:  y = -3(2) 2 + 12(2) + 29  = -3(4) + 24 + 29  = -12 + 24 + 29  = 41  Therefore, the maximum y-value is 41.

3.  Express the quadratic relation y=2x 2 +8x+5 in vertex form. Then sketch the relation by hand.  We can’t factor the equation, so find the axis of symmetry, since vertex lies on it ◦ We just need 2 points with the same y-value, and find the x-value in between ◦ Let’s try y = 5

4.  5 = 2x 2 +8x+5  0 = 2x 2 + 8x  0 = 2x(x + 4)  Either x = 0 or x = -4, so x = -2 is the axis of symmetry.  The x-coordinate of the vertex is x = -2  To find the y-value, plug back into the equation:  y = 2x 2 + 8x + 5  = 2(-2) 2 + 8(-2) + 5  = 2(4) + (-16) + 5  = 8 – 16 + 5  -3  Thus, the vertex of the parabola is (-2, -3)  The equation of the parabola in vertex form is y= 2(x+2) 2 – 3 (the value of “a” is the same in standard and vertex form)

5.  If a quadratic relation is in standard form and cannot be factored fully, you can use partial factoring to help you determine the axis of symmetry to determine the coordinates of the vertex