How do you find the important points to graph #y = -x^2 + 3#?

Question

1928 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-02T05:45:43

vertex # (0, 3)#
#x#-intercepts: #(-sqrt(3), 0), (sqrt(3), 0)#
#y#-intercept: #(0, 3)#

Given: #y = -x^2 + 3#

With the equation in the form: #Ax^2 + Bx + C = 0#,

the vertex is at #(-B/(2A), f(-B/(2A)))#,

the axis of symmetry is #x = -B/(2A)#

If the coefficient #A < 0#, the vertex is a maximum

If the coefficient #A > 0#, the vertex is a minimum

For the given equation:

#-B/(2A) = 0/-2 = 0#

#f(0) = -(0)^2 + 3 = 3#

vertex # (0, 3)# is a maximum; axis of symmetry: #x = 0#

#color(blue) ("Find x-intercepts")# by setting #y = 0#:

#0 = -x^2 + 3#

#-3 = -x^2#

#x^2 = 3 => x = +- sqrt(3)#

#x#-intercepts: #(-sqrt(3), 0), (sqrt(3), 0)#

#color(red) ("Find y-intercept")# by setting #x = 0#:

#y = -(0)^2 + 3 => y = 3#

#y#-intercept: #(0, 3)#, which is the vertex