Suppose f(x) is a quadratic function with only one root at x=−5 and y-intercept (0,75). Write the equation for this function.

Answer 1

Quadratics (2nd order polynomials) always have 2 total zeros, either imaginary or real, but since imaginary roots only come in conjugate pairs, the only possibility is to have 2 imaginary zeros or two real zeros. So, when someone says a quadratic only has one zero, it MUST be that that zero is a double (two of the same number).

So, we can write the quadratic as f(x) = (x + 5)(x + 5) BUT, to be complete, we must add a multiplier in front of the factors. That makes our quadratic:

f(x) = a(x + 5)(x + 5) where "a" is the multiplier. To find out the value of "a", use the fact that f(0) = 75:

75 = a(0 + 5)(0 + 5)

75 = 25a

a = 3

Therefore the quadratic is:

f(x) = 3(x + 5)(x + 5)

f(x) = 3(x² + 10x + 25)

f(x) = 3x² + 30x + 75)

Answer 2

y = a(x+5)^2

75 = a(5^2) = 25a

a = 75/25 = 3

y = 3(x+5)^2

= 3x^2 +30x +75