How do you find the vertex and the intercepts for #y = 3x^2 - 18x + 4#?

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Answer 1 · 2017-05-29T10:58:51

Vertex is at #(3,-23)# , y-intercept is at #(0,4)# and
x-intercepts are at #(0.23,0) and (5.77,0)#

#y=3x^2-18x+4 = 3(x^2-6x)+4 = 3(x^2-6x+9) -27 +4 # or
#y=3(x-3)^2 -23 # Comparing with vertex form of equation

#y=a(x-h)^2+k ; (h,k)# being vertex we find here # h=3 ; k=-23#

So vertex is at #(3,-23)# , y-intercept can be found by putting #x=0# in the equation #:.y=4 ;# y-intercept is at #(0,4)#

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

#:. 3(x-3)^2 -23 =0 or 3(x-3)^2 = 23 or (x-3)^2 =23/3 # or

# (x-3) = +- sqrt(23/3) or x = 3 +- sqrt(23/3) :. x ~~ 5.77(2dp) and x ~~0.23(2dp)#

x-intercepts are at #(0.23,0) and (5.77,0)#
graph{3x^2-18x+4 [-80, 80, -40, 40]} [Ans]