How do you find the vertex and intercepts for #y=-x^2+4x+12#?

1 Answer
Nghi N
Jul 9, 2017

#y = - x^2 + 4x + 12#
x-coordinate of vertex;
#x = - b/2a = -4/-2 = 2#
y-coordinate of vertex:
#y(2) = - 4 + 8 + 12 = 16#
Vertex (2, 16)
Make x = 0 --> y-intercept = 12
Make y = 0, and solve the quadratic equation:
#y = - x^2 + 4x + 12 = 0#
Find 2 real roots knowing the sum (b = 4) and product (ac = - 12). The 2 x-intercepts are: x = - 2 and x = 6.
graph{- x^2 + 4x + 12 [-40, 40, -20, 20]}

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