If the zeros of a quadratic function f are -3 and 5 what is the value of a+b+c

Answer 1

There is nothing in the statement of the problem that says the quadratic is monic (leading coefficient is 1). So the most we can conclude is that a + b + c = -16a. This follows from writing the function as f(x) = a(x+3)(x-5), expanding, and adding the coefficients.

Answer 2

If the zeros of a quadratic function (a.k.a.: a parabola) are x = -3 and x = 5, that means that the graph of the parabola has x-intercepts (crosses the x-axis) at x = -3 and x = 5.

Working backward, we can then say that if the equation of the quadratic function were in factored form, it would look like this:

y = (x + 3)(x - 5) *remember that the signs of the x-intercepts are switched in factored form!

In order to find the values of a, b, and c, we will need to first convert the quadratic function from factored form into standard form: ax² + bx + c. We can do this easily by F.O.I.L.-ing:

y = (x + 3)(x - 5)

= x² - 5x + 3x -15 Combining like terms, we get:

y = x² - 2x - 15 Standard Form

We can now see that a = 1, b = -2, and c = - 15.

Finally, to find the value of a + b + c, we just add them: 1 + (-2) + (-15) = -16

Answer 3

(x + 3)(x - 5)

a = 1

b = -2

c = -15

-16