A quadratic equation of type y=ax^2+bx+cy=ax2+bx+c, which is equation of a parabola, can be written in vertex form i.e.
y=a(x-h)^2+ky=a(x−h)2+k
For example, let us have the equation y=-3x^2-6x+9y=−3x2−6x+9, them we can write it as
y=-3(x^2+2x+1)+3+9=-3(x+1)^2+12y=−3(x2+2x+1)+3+9=−3(x+1)2+12
Similarly y=2x^2+3x-7=2(x^2+3/2x)-7y=2x2+3x−7=2(x2+32x)−7
= 2(x^2+2xx3/4x+(3/4)^2)-2xx(3/4)^2-72(x2+2×34x+(34)2)−2×(34)2−7
= 2(x+3/4)^2-65/8=2(x+3/4)^2-8 1/82(x+34)2−658=2(x+34)2−818
Observe that (x-h)^2>=0(x−h)2≥0 and hence
if a>0a>0, a(x-h)^2a(x−h)2 is positive and the minimum value of yy, when x=hx=h is kk i.e. the parabola rises on either side and has lowest point as (h,k)(h,k). For example, y=2x^2+3x-7=2(x+3/4)^2-8 1/8y=2x2+3x−7=2(x+34)2−818 and graph of y=2x^2+3x-7y=2x2+3x−7 has minima at (-3/4,8 1/8)(−34,818) and observe that (-3/4,8 1/8)(−34,818) appears as a vertex.
graph{2x^2+3x-7 [-4, 4, -11.36, 8.64]}
but if a<0a<0, a(x-h)^2a(x−h)2 is negative and the maximum value of yy, when x=hx=h is kk i.e. the parabola falls on either side and has highest point as (h,k)(h,k). For example, y=-3x^2-6x+9=-3(x+1)^2+12y=−3x2−6x+9=−3(x+1)2+12 and graph of y=-3x^2-6x+9y=−3x2−6x+9 has maxima at (-1,12)(−1,12) and observe that (-1,12)(−1,12) appears as a vertex.
graph{-3x^2-6x+9 [-5.433, 2.567, -6, 14]}
