How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y=x^2+2x-24#?

1 Answer
Nghi N.
Jun 8, 2015

x of vertex: #x = (-b/(2a)) = -4/2 = -2#

y of vertex: y = f(-2) = 4 -4 - 24 = -24

x of axis of symmetry = x of vertex = -2.

To find x-intercepts, make #y = x^2 + 2x - 24 = 0. #Roots have different signs. Compose factor pairs of (-24) -> (-2, 12)(-4, 6). This sum is (6 - 4 = 2 = b). Then the 2 real roots are the opposites: 4 and -6.
two x-intercepts: #4 and -6#

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