Answer:
(3, 0) and (5, 0)
Step-by-step explanation:
we have
[tex]y=x^{2}-8x+15[/tex]
we know that
The x-intercepts are the values of x when the value of y is equal to zero
so
For y=0
[tex]x^{2}-8x+15=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-8x+15=0[/tex]
so
[tex]a=1\\b=-8\\c=15[/tex]
substitute in the formula
[tex]x=\frac{-(-8)\pm\sqrt{-8^{2}-4(1)(15)}} {2(1)}[/tex]
[tex]x=\frac{8\pm\sqrt{4}} {2}[/tex]
[tex]x=\frac{8\pm2} {2}[/tex]
[tex]x=\frac{8+2} {2}=5[/tex]
[tex]x=\frac{8-2} {2}=3[/tex]
so
x=3, x=5
therefore
The x-intercepts are (3,0) and (5,0)
