Question

what is the line of symmetry for the parabola whose equation is y=-x^2-x+2

Answer 1

Answer:

$\huge \boxed{\sf \ \ x=-\dfrac{1}{2} \ \ }$

Step-by-step explanation:

Hello,

We need to express this parabola using this kind of expression

   $y=a(x-b)^2+c$

and then, the line of symmetry will be the line x = b

Let's do it !

$\text{*** Complete the square ***} \\ \\ x^2+x=x+2\cdot \dfrac{1}{2}\cdot x=(x+\dfrac{1}{2})^2-\dfrac{1^2}{2^2}=(x+\dfrac{1}{2})^2-\dfrac{1^2}{4} \\ \\ \text{*** Apply it to our parabola } \\ \\y=-x^2-x+2=-(x^2+x)+2=-[(x+\dfrac{1}{2})^2-\dfrac{1}{4}]+2 = -(x+\dfrac{1}{2})^2+\dfrac{1+2*4}{4}= -(x+\dfrac{1}{2})^2+\dfrac{9}{4} \\ \\ \text{*** It comes ***} \\ \\ \Large \boxed{\sf \ \ y=-(x+\dfrac{1}{2})^2+\dfrac{9}{4} \ \ }$

So the line of symmetry is

$\huge \boxed{\sf \ \ x=-\dfrac{1}{2} \ \ }$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you