Given:
#f(x)=-x^2+4x+4# is a quadratic equation in standard form:
#f(x)=ax^2+bx+c#,
where:
#a=-1#, #b=4#, and #c=4#.
Vertex: maximum or minimum point of a parabola
The #x# value of the vertex is the same as the axis of symmetry that divides the parabola into two equal halves.
#x=(-b)/(2a)#
#x=(-4)/(2*-1)#
#x=(-4)/(-2)# #larr# Two negatives make a positive.
#x=2#
To determine the #y# value of the vertex, substitute #y# for #f(x)#. Substitute #2# for #x# and solve for #y#.
#y=-(2)^2+4(2)+4#
Simplify.
#y=-4+8+4#
#y=8#
The vertex is #(2,8)#.
Since #a<0#, the vertex is the maximum point and the parabola opens downward.
X-Intercepts: value of #x# when #y=0#
Substitute #0# for #f(x)# and solve for #x# using the quadratic formula.
#0=-x^2+4x+4#
Quadratic formula
#x=(-b+-sqrt(b^2-4ac))/(2a)#
Plug in the known values.
#x=(-4+-sqrt((4)^2-4*-1*4))/(2*-1)#
Simplify.
#x=(-4+-sqrt(16+16))/(-2)#
Simplify #16+16# to #32#.
#x=(-4+-sqrt32)/(-2)#
Prime factorize #32#.
#x=(-4+-sqrt((2xx2)xx(2xx2)xx2))/(-2)#
Simplify.
#x=(-4+-2xx2sqrt2)/(-2)#
#x=(-4+-4sqrt2)/(-2)#
Factor out the common #2#.
#x=(color(red)cancel(color(black)(-4))^2+-color(red)cancel(color(black)(4))^2sqrt2)/(color(red)cancel(color(black)(-2))^1#
#x=(2+-2sqrt2)#
Solutions for #x#.
#x=2+2sqrt2,##2-2sqrt2#
Simplify.
#x=2(1+sqrt2),##2(1-sqrt2)#
x-intercepts: #(2(1+sqrt2)),0),# #(2(1-sqrt2)),0)#
Approximate intercepts: #(-0.828,0),##(4.83,0)#
Y-Intercept: value of #y# when #x=0#.
Substitute #0# for #x# and solve for #y#.
#y=-(0^2)+4(0)+4#
#y=4#
y-intercept: #(0,4)#
Plot the points and sketch a parabola through the dots. Do not connect the dots.
graph{y=-x^2+4x+4 [-16.02, 16.02, -8.01, 8.01]}
