we have
\(f(x)=x^{2} +2x-15\)
we know that
The x-intercept is the value of x when the value of the function is equal to zero
so
in this problem
Find the roots of the function
equate the function to zero
\(x^{2} +2x-15=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(x^{2} +2x=15\)
Complete the square. Remember to balance the equation by adding the same constants to each side
\(x^{2} +2x+1=15+1\)
\(x^{2} +2x+1=16\)
Rewrite as perfect squares
\((x+1)^{2}=16\)
Square root both sides
\((x+1)=(+/-)\sqrt{16}\)
\((x+1)=(+/-)4\)
\(x=-1(+/-)4\)
\(x1=-1+4=3\)
\(x2=-1-4=-5\)
\(x^{2} +2x-15=(x-3)(x+5)\)
therefore
the answer is
the x-intercepts are the points \((3,0)\) and \((-5,0)\)
