f(2) = 19 means that
4a+2b+c = 19
f(-1) = 8 means that
a-b+c = 8
f'(-1) = 0 means that
-2a+b = 0
Solve all of those as usual for a,b,c and you get
f(x) = 11/9 x^2 + 22/9 x + 83/9
or, skipping the calculus, you have the vertex, so you know that
f(x) = a(x+1)^2 + 8
so
9a+8 = 19
a = 11/9
so f(x) = 11/9 (x+1)^2 + 8
Determine a quadratic function f(x)= ax^2 +bx + c if its graph passes through the point (2,19) and has a horizontal tangent at (-1,8).
Thank you!!
1 answer