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

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) (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 use
y-intercept is
if

V = (-(1/2),f(-1/2))=(-(1/2),(1/4)-(1/2)-2)=(-(1/2),-(9/4))=(-0.5,-2.25)
x-intercepts:
=
=
so x-intercepts are -2 and 1
y-intercept is f(0)=-2

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