How do you find the vertex and intercepts for f(x) = 2x^2 - 4x + 3?
1 Answer
Jun 14, 2018
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]}