If the function f(x)=ax^2+bx+c has x-intercepts (1,0) and (-5,0) and y-intercept (0,-1) Then f(3) =?

Question

If the function f(x)=ax^2+bx+c has x-intercepts (1,0) and (-5,0) and y-intercept (0,-1) Then f(3) =

Answer 1

0 = a*1^2 + b*1 + c ... eq#1

0 = a*(-5)^2 + b*(-5) + c ... eq#2

-1 = a*0^2 + b*0 + c ... eq#3

c = -1 ... from eq#3

0 = a + b - 1 ... from eq#1

0 = 25a - 5b - 1 ... from eq#2

0 = 5a + 5b - 5 ... from eq#1

0 = 30a - 6 ... add eqs

a = 6/30 = 1/5

0 = 1/5 + b - 1 ... from eq#1

b = 4/5

f(x) = x^2/5 + 4x/5 - 1

f(3) = 3^2/5 + 4*3/5 - 1

f(3) = 16/5

Answer 2

f(1) = a + b + c = 0

f(-5) = 25a - 5b + c = 0

f(0) = c = -1

Solve for a, b, c and plug in 3 for x

a + b = 1

25a - 5b = 1

b = 1 - a

25a - 5 + 5a = 1

30a = 6

a = 1/5

b = 4/5

c = -1

f(3) = 1/5 * 9 + 4/5 * 3 - 1 = 9/5 + 12/5 - 5/5 = 16/5