SOLUTION: I have seen how to solve these equations when solving for y where you make an x any y table start with -2 to 2 and input the numbers to solve the parabola. How do you use that same

Algebra ->  Linear-equations -> SOLUTION: I have seen how to solve these equations when solving for y where you make an x any y table start with -2 to 2 and input the numbers to solve the parabola. How do you use that same      Log On


   

Question 155444: I have seen how to solve these equations when solving for y where you make an x any y table start with -2 to 2 and input the numbers to solve the parabola. How do you use that same process for a problem like this?
Find the vertex and intercepts of the function f(x)= x^2 + x - 2

Answer by oscargut(2103) About Me  (Show Source):
You can put this solution on YOUR website!
if you have f(x)=ax^2+bx +c
vertex is V=(-b/a,f(-b/a))
x-intercepts are the roots of f(x)
you have to usex+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
y-intercept is f%280%29=c
if
f%28x%29=+x%5E2+%2B+x+-+2
V = (-(1/2),f(-1/2))=(-(1/2),(1/4)-(1/2)-2)=(-(1/2),-(9/4))=(-0.5,-2.25)
x-intercepts:
x+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-2%29+%29%29%2F%282%2A1%29+=
x+=+%28-1+%2B-+sqrt%289%29%29%2F2+=x+=+%28-1+%2B-+3%29%2F2+
so x-intercepts are -2 and 1
y-intercept is f(0)=-2