How do you find the vertex and intercepts for #f(x) = 2x^2 - 4x + 3#?

1 Answer
Jun 14, 2018

#"see explanation"#

Explanation:

#"express the quadratic in "color(blue)"vertex form"#

#•color(white)(x)y=a(x-h)^2+k#

#"where "(h,k)" are the coordinates of the vertex and a is"#
#"a multiplier"#

#"using the method of "color(blue)"completing the square"#

#f(x)=2(x^2-2x+3/2)#

#color(white)(f(x))=2(x^2+2(-1)x+1-1+3/2)#

#color(white)(f(x))=2(x-1)^2+1#

#"vertex "=(1,1)#

#"for y-intercept set x = 0"#

#f(0)=3rArr(0,9)#

#"for x-intercepts set y = 0"#

#2(x-1)^2+1=0#

#(x-1)^2=-1/2#

#"this has no real solutions hence no x-intercepts"#
graph{2x^2-4x+3 [-10, 10, -5, 5]}