Question

Find the axis of symmetry of $y=x^2-x-12$.

• A. $x=0.5$
• B. $x=5$
• C. $x=1.5$
• D. None of these

Answer 1

The correct option is A $x=0.5$

For the quadratic function $y=a{x}^2+bx+c$, the coordinate of the vertex are $(\frac{-b}{2a},\frac{-D}{4a})$ where $D=b^2-4ac$

Comparing $y=a{x}^2+bx+c$ with $y=x^2-x-12$

we get $a=1,b=-1,c=-12$

Now, putting the values of $a,b,c$ we get

$(\frac{-b}{2a},\frac{-D}{4a})=(\frac{1}{2},\frac{-49}{4})$

$Axis$ $of$ $Symmetry$ passes through the Vertex of this Parabola $Perpendicular$ to the $x-axis$ and therefore its Equation is

$x=\frac{-b}{2a}=\frac{1}{2}$