Find constants a, b and c such that f(t)=at^2+bt+c is a quadratic function that has horizontal intercepts at t=−3 and t=4 and a graph containing the point (1,−3)

Question

To my understanding my x-int are (-3,0) and (4,0) and y-int (1,-3). I'm not to confident on what to do next but I assume you need to find a in f(t)=at^2+bt+c?

Answer 1

x intercepts -3 and 4 means an axis of symmetry t=(-3+4)/2 = 1/2

1/2 is the x or t coordinate of the vertex

f(t) = y = (t-1/2)^2 + k

-3 = (1-1/2)^2 +k = 3/4 +k

k = -3-3/4 = -15/4

f(t) = (t-1/2)^2 -15/4

= t^2 -t + 1/4 -15/4

= t^2 +t - 14/4

= t^2 +t -7/2

a =1, b= 1, c= -7/2

for

f(t) = at^2 +bt + c