SOLUTION: f(x)=x^2+x-2 put the equation in vertex form determine the vertex point deter,ine the x-intercept the y-intercept, and the axis of symmetry

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: f(x)=x^2+x-2 put the equation in vertex form determine the vertex point deter,ine the x-intercept the y-intercept, and the axis of symmetry       Log On


   

Question 1199204: f(x)=x^2+x-2
put the equation in vertex form
determine the vertex point
deter,ine the x-intercept the y-intercept, and the axis of symmetry
Found 2 solutions by MathLover1, josgarithmetic:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=x%5E2%2Bx-2
put the equation in vertex form: f%28x%29=%28x-h%29%5E2%2Bk
f%28x%29=x%5E2%2Bx-2...complete the square
f%28x%29=%28x%5E2%2Bx%2Bb%5E2%29-b%5E2-2......b=1%2F2
f%28x%29=%28x%5E2%2Bx%2B%281%2F2%29%5E2%29-%281%2F2%29%5E2-2
f%28x%29=%28x%2B1%2F2%29%5E2-1%2F4-2
f%28x%29=%28x%2B1%2F2%29%5E2-9%2F4


determine the vertex point
=>h=-1%2F2,k=-9%2F4
the vertex point is:(-1%2F2,-9%2F4)
the x-intercept : set f%28x%29=0
0=%28x%2B1%2F2%29%5E2-9%2F4
%28x%2B1%2F2%29%5E2=9%2F4
sqrt%28%28x%2B1%2F2%29%5E2%29=sqrt%289%2F4%29
x%2B1%2F2=3%2F2=>x=3%2F2-1%2F2=>x=1
or
-%28x%2B1%2F2%29=3%2F2=>-x-1%2F2=3%2F2=>-intercept : set x=0
f%280%29=%280%2B1%2F2%29%5E2-9%2F4
f%280%29=1%2F4-9%2F4
f%280%29=8%2F4
f%280%29=2
and the axis of symmetry is:x=-1%2F2



Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=x%5E2%2Bx-2
Term to Complete The Square, %281%2F2%29%5E2=1%2F4, based on coefficient for the x term;
f%28x%29=x%5E2%2Bx%2B1%2F4-1%2F4-2
%28x%5E2%2Bx%2B1%2F4%29-1%2F4-8%2F4
highlight_green%28f%28x%29=%28x%2B1%2F2%29%5E2-9%2F4%29

Vertex, ( -1/2, -9/4 )

To find the x and y intercepts
%28x%2B1%2F2%29%5E2-9%2F4=0
%28x%2B1%2F2%29%5E2=9%2F4
x%2B1%2F2=0%2B-+3%2F2
x=-1%2F2%2B-+3%2F2---------the x-intercepts
-
y=%280%2B1%2F2%29%5E2-9%2F4
y=1%2F4-9%2F4
y=-8%2F4=-2------------the y-intercept