How do you find the axis of symmetry, vertex and x intercepts for #y=-x^2+3x-3#?

1 Answer
Dec 16, 2017

Axis of symmetry: #x=3/2#

Vertex: #(3/2,-3/4)#

Complex solutions: #x=(3+sqrt(3)i)/(2),##(3-sqrt(3)i)/(2)#

Explanation:

Given:

#y=-x^2+3x-3# is a quadratic equation in standard form:

#ax^2+bx+c#,

where:

#a=-1#, #b=3#, and #c=-3#

Axis of symmetry: vertical line that divides the parabola into two equal halves, and is also the #x#-value of the vertex. For a quadratic equation, #x=(-b)/(2a)#.

#x=(-3)/(2*(-1))#

#x=(-3)/-2#

#x=3/2# #larr# Axis of symmetry

Vertex: the maximum or minimum point #(x,y)# of a parabola. Since #a<0#, the vertex will be the maximum point and the parabola opens downward.

Substitute #3/2# for #x# in the equation and solve for #y#.

#y=-(3/2)^2+3(3/2)-3#

Simplilfy.

#y=-9/4+9/2-3#

Multiply each number by a fraction equal to #1# so that each of their denominators is #4#. For example, #6/6=1#. Recall that a whole number is understood to have a denominator of #1#.

#y=-9/4+9/2xxcolor(teal)(2/2)-3/1xxcolor(magenta)(4/4#

Simplify.

#y=-9/4+18/4-12/4#

#y=(-9+18-12)/4#

#y=-3/4#`

Vertex: #(3/2,-3/4)# or #(1.5,-0.75)#

X-intercepts: The parabola does not cross the x-axis, so there are no real solutions, so there are no intercepts, but there are complex solutions.

Substitute #0# for #y#. Use the quadratic formula to solve for #x#.

#0=-x^2+3x-3#

Quadratic Formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the known values.

#x=(-3+-sqrt(3^2-4*-1*-3))/(2*-1)#

Simplify.

#x=(-3+-sqrt(-3))/(-2)#

Complex solutions for #x#.

#x=(3+sqrt(3)i)/(2),##(3-sqrt(3)i)/(2)#

graph{y=-x^2+3x-3 [-16.08, 15.94, -13.46, 2.56]}