How do you find the vertex and the intercepts for #y= -x^2 + 9#?

1 Answer
Feb 9, 2017

Vertex: #(0,9)#
Y-intercept: #y=9#
X-intercepts: #x=-3 and x=+3#

Explanation:

Remember the standard vertex form for a parabola is
#color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b#
with vertex at #(color(red)a,color(blue)b)#

Re-writing the given equation: #y=-x^2+9#
into explicit standard vertex form:
#color(white)("XXX")y=color(green)(-1)(x-color(red)0)^2+color(blue)9#
with vertex at #(color(red)0,color(blue)9)#

The Y-intercept is the value of #y# when #x=color(magenta)0#
#color(white)("XXX")y=-color(magenta)0^2+9=9#

The X-intercepts are the values of #x# possible when #y=color(orange)0#
#color(white)("XXX")color(orange)0=-x^2+9#
#color(white)("XXX")rarr x^2=9#
#color(white)("XXX")rarr x=+-3#

Here is the graph for verification purposes:
enter image source here