How do you graph the quadratic function and identify the vertex and axis of symmetry for #y=-x^2+4x-2#?

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Answer 1 · 2017-03-21T13:59:12

Vertex#->(x,y)=(2,2)#
Axis of symmetry #->x=2#

As the #x^2# part is negative the graph is of general shape #nn#

The axis of symmetry coincides with #x_("vertex")#

#color(blue)("Determine the vertex and axis of symmetry")#

Write as #y=-1(xcolor(red)(-4)x)-2#

#"Axis of symmetry "=x_("vertex")=(-1/2)xx(color(red)(-4)) = +2#

By substitution:

#y_("vertex")=-(2)^2+4(2)-2 = +2#

Vertex#->(x,y)=(2,2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the y-intercept")#

Consider the given equation:#" "=-x^2+4xcolor(magenta)(-2)#

#y_("intercept")=color(magenta)(-2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the x-intercept")#

Using #y=ax^2+bx+c" where "x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=-1"; "b=+4"; "c=-2#

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

I will let you finish this bit.

Tony B