How do you find the vertex and the intercepts for #y = (x+4)(x-6) #?

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Answer 1 · 2017-08-18T11:01:00

Vertex is at #(1,-25)# , #x# intercepts are at #(-4,0) **and** (6,0)# and #y# intercept is at #(0,-24)#

#y = (x+4)(x-6) or y= x^2 -2x -24 # or

#y = x^2 -2x +1 -1 -24 or y = (x-1)^2 -25 #

Comparing with vertex form of equation #y=a(x-h)^2+k ; (h,k)#

being vertex , here #h=1 , k = - 25 # , So vertex is at #(1,-25)#

To find #y# intercept , putting #x=0# in the equation we get

#y = (x+4)(x-6) or y = (0+4)(0-6) = -24 # ,

#y# intercept is at #y =-24 # or at #(0,-24)# . To find #x# intercept ,

putting #y=0# in the equation we get , #0 = (x+4)(x-6) #

#:. x =-4 , x = 6 :. x# intercepts are at #x =-4 , x =6 #or at

#(-4,0) and (6,0)#.

graph{(x+4)(x-6) [-80, 80, -40, 40]} [Ans]