Question

Identify the vertex and axis of symmetry of the quadratic function.f(x)= - 3(x + 2)^2 + 5

Answer 1

We are given the equation of a parabola, and we are asked to find its vertex and axis of symmetry

The general form of a quadratic equation is the following

$f(x)=a(x-h)^2+k$

Written in this form, the point (h,k) represent the vertex of the parabola, that means that in the equation given by the problem

$f(x)=-3(x+2)^2+5$

The vertex is the point

$(h,k)=(-2,5)$

To find the axis of symmetry we need to expand the equation by solving the parenthesis and simplifying

$f(x)=-3(x^2+4x+4)+5$$f(x)=-3x^2-12x-12+5$$f(x)=-3x^2-12x-7$

Written in this form, we have that the axis of symmetry is given by

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

where the coeficient b and a, come from the equation, knowing that the general form is the following

$f(x)=ax^2+bx+c$

therefor, the axis of symmetry is equal to

$x=-\frac{-12}{2(-3)}=-2$