Hello!The answer is: x intercepts at -3 and -5.Why?To find the x-intercepts we just need to find where the function tends to 0,So,[tex]x^{2}+8x+15=0[/tex][tex]a=1\\b=8\\c=15[/tex]We can solve this quadratic equation by finding two numbers which its sum give as result 8 and multiplied each other gives as result 15, that numbers would be 3 and 5, knowing that we can rewrite the quadratic equation by the following way:[tex]x^{2} + 8x + 15= (x+3)*(x+5)[/tex]For the new equation, we just need to find the values of x that make it 0,When x is -3[tex](-3+3)*(-3+5)=0*-2=0[/tex]When x is -5[tex](-5+3)*(-5+5)=-2*0=0[/tex]So, the intercepts of the given function are: -3 and -5We can also find the x-intercepts using the quadractic formula:[tex]\frac{-b+-\sqrt{b^{2} -4*a*c}}{2a}[/tex]By substituting we have:[tex]\frac{-8+-\sqrt{8^{2} -4*1*15}}{2*1}=\frac{-8+-\sqrt{64-60}}{2}=\frac{-8+-\sqrt{4}}{2}=\frac{-8+-2}{2} \\\\x1=\frac{-8+2}{2}=-3\\\\x2=\frac{-8-2}{2}=-5[/tex]So, the x- intercepts of the given function are: -3 and -5Have a nice day!