Answer:
The zeros of the function are x = -9 and x = 1.
Step-by-step explanation:
The zeros of a function f(x) are the values of x for which f(x) = 0.
In this problem:
[tex]f(x) = (x+4)^{2} - 25[/tex]
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = (x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
So
[tex]f(x) = x^{2} + 8x + 16 - 25[/tex]
[tex]f(x) = x^{2} + 8x - 9[/tex]
Zeros
[tex]x^{2} + 8x - 9 = 0[/tex]
This means that [tex]a = 1, b = 8, c = -9[/tex]
[tex]\bigtriangleup = b^{2} - 4ac = (8)^{2} - 4*1(-9) = 100[/tex]
[tex]x_{1} = \frac{-8 + \sqrt{100}}{2*1} = 1[/tex]
[tex]x_{2} = \frac{-8 - \sqrt{100}}{2*1} = -9[/tex]
The zeros of the function are x = -9 and x = 1.
