Question 254208
vertex is (-3, -2) and it passes through the point (-2, 0)
All parabolas are symmetric, so the other point would be (-4,0).
We can create a quadratic model based on this using
{{{Y = aX^2 + bX + c}}}
(i) {{{-2 = 9a + -3b + c}}}
(ii) {{{0 = 4a -2b + c}}}
(iii) {{{0 = 16a -4b + c}}}
eliminate the c from all 3 equations (i) -(ii) to get
(iv) {{{-2 = 5a - b}}}
and (iii) - (ii) to get
(v) {{{0 = 12a -2b}}}
multiply (iv) by -2 to get
(vi) {{{4 = -10a + 2b}}}
add (v) and (vi) to get
(vii) {{{4 = 2a}}}
* a = 2.
Sub -2 into a for (v) to get
* b = 12
Sub values of a and b into (i) to get
* c = 16
We get
 y = 2x^2 + 12x + 16