How do you find the intercepts, vertex and graph #f(x)=-3x^2-4x#?

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Answer 1 · 2018-05-30T15:48:53

See graph.

Set #y=0# solve for #x# by factoring to find the x-intercept(s) if they exist (remember #f(x) = y#):

#f(x)=-3x^2-4x#

#0=-3x^2-4x#

#0=x(-3x-4)#

#x=0# and #x=-4/3#

Set #x=0# solve for #y# to find the y-intercept:

#f(x)=-3*0^2-4*0 = 0#

#y = 0#

in the form:

#ax^2 + bx +c# the formula for axis of symmetry is:

#aos = (-b)/(2a)#

#f(x)=-3x^2-4x +0#

a=-3
b=-4

#aos = (-(-4))/(2*-3)#

#aos = -2/3#

The formula for the vertex is:

vertex #(x,y) = (aso, f(aos))#

#f(-2/3)=-3(-2/3)^2-4*(-2/3)+0#

#f(-2/3)=4/3#

vertex = #(-2/3, 4/3)

Finally since #a<0# the Parabola opens down and has a maximum at the vertex:

graph{-3x^2-4x [-5, 5, -2.5, 2.5]}