How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=-3x^2-4x#?

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Answer 1 · 2017-07-17T12:15:40

Vertex is at #(2/3, 4/3)#, y-intercept is at #(0,0)# and
axis of symmetry is #x=2/3#

#f(x) = -3x^2+4x or f(x) = -3(x^2-4/3x)# or

#f(x) = -3(x^2-4/3x +(2/3)^2) + 4/3# or

#f(x) = -3(x-2/3)^2 + 4/3# comparing with vertex form of

equation #y= a(x-h)^2+k ; (h,k) # being vertex we get here

#h=2/3, k= 4/3# So vertex is at #(2/3, 4/3)#

y-intercept can be found by putting #x=0# in the equation

#f(x) = -3x^2+4x :. f(x) =0 :.# y-intercept is at #(0,0)#

Axis of symmetry is #x=2/3#

graph{-3x^2+4x [-10, 10, -5, 5]} [Ans]