Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of
(- 3, - 25), and x intercepts at x = -8 and x = 2. (Do not include the negative sign in your answer.)
y = x2 + 6x -

Question

Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of 
(- 3, - 25), and x intercepts at x = -8 and x = 2. (Do not include the negative sign in your answer.)
y = x2 + 6x -

campbell_st · Answer

well using the zeros the polynomial is and a is a constant

y = a(x +8)(x - 2)

so 

y = a(x^2 + 6x - 16)

you now need to find the value of a by substituting the vertex  x= -3, y = -25

$$-25 = a((-3)^2 + 6(-3) - 16) $$

-25 = a(9 - 18 - 16)

a = 1

so the equation is 

$$y = x^2 + 6x - 16$$