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 -5
We 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 -5
Have a nice day!
