How do you find the vertex and the intercepts for #f(x)=x^2-x-12 #?

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Answer 1 · 2018-05-08T12:32:44

Vertex is at #(0.5, -12.25) #, x intercepts are at
#(-3,0) and(4,0)# and y intercept is at #(0,-12)#

# f(x)= y= x^2-x-12 or y= (x^2-x +1/4)- 49/4# or

# y= (x -0.5)^2 - 12.25# Comparing with vertex form of

equation #f(x) = a(x-h)^2+k ; (h,k)# being vertex we find

here #h=0.5 , k=-12.25 :.# Vertex is at #(0.5, -12.25) # .

y intercept is found by putting #x=0# in the equation

#y = x^2- x-12 :. y = -12# or at #(0,-12)#

y intercept is at #(0,-12)#

x intercept is found by putting #y=0# in the equation

#y = x^2- x-12 or x^2- x-12 =0 # or

# x^2- 4 x +3 x-12 =0 or x(x-4) +3 (x-4) =0 # or

#(x-4)(x+3) =0 :. x = -3 and x = 4#

x intercepts are at #(-3,0) and (4,0)#

graph{x^2-x-12 [-40, 40, -20, 20]} [Ans]