Step 1
Given;
[tex]f(x)=x^2-2x-15[/tex]Required; To find the axis of symmetry
Step 2
[tex]\begin{gathered} \mathrm{For\:a\:parabola\:in\:standard\:form}\:y=ax^2+bx+c \\ \mathrm{the\:axis\:of\:symmetry\:is\:the\:vertical\:line\:that\:goes\:through\:the\:vertex}\:x=\frac{-b}{2a} \\ a=1,b=-2 \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-\left(-2\right)}{2\cdot \:1} \\ Simplify \\ x=1 \end{gathered}[/tex]Answer;
[tex]\begin{gathered} The\text{ axis of symmetry of f\lparen x\rparen=x}^2-2x-15\text{ is;} \\ x=1 \end{gathered}[/tex]
