Question

What are the zeros of the function f(x)=-x^2+13x-36

Answer 1

For this case we have the following function:

$f (x) = - x ^ 2 + 13x-36$

To find the zeros of the function we make $y = 0$and solve for "x", then:

$0 = -x ^ 2 + 13x-36$

We multiply by -1 on both sides of the equation:

$0 = x ^ 2-13x + 36$

We factor the equation, for this we look for two numbers that, when multiplied, result in 36 and when added, result in -13. These numbers are -9 and -4.

$(-9) * (- 4) = 36\\-9-4 = -13$

Thus, the factored equation is:

$(x-9) (x-4) = 0$

Therefore, the roots are:

$x_ {1} = 9\\x_ {2} = 4$

Answer:

$x_ {1} = 9\\x_ {2} = 4$