Question 144449: I barely even know how to find the quadratic formula of f(x), i don't even know if i can find the roots of this given function, and then graph the function of g(x) = x2 - x - 12. What do i do just to get started on this someone. Any help would be appreciated. Thank you in advance.
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! This will be a pseudo-lesson of factoring and graphing. Do not become afraid of all the letters I use in the general case...
Factor first:
x^2-x-12
If we can factor a quadratic it will generally be in form:
(ax+b)(cx+d)
Now if we apply FOIL to that:
acx^2+dax+bcx+bd
=acx^2+(da+bc)x+bd
As you can see, the number in front of the x term is the sum of da and bc. On the other hand, the "loose" term, that is, the one without an x, is the product bd.
So we want -1=da+bc and -12=bd.
We can achieve this by listing the factors of -12 and seeing which add to -1 (when one is made negative).
12=1*2*2*3 which means 1,2,3,4,6,and 12 are factors. Now, we can see that -4+3=-1, and -4*3=-12. These have to be the two factors!
So, x^2-x-12=(x-4)(x+3).
Now, to find the zeros/solutions, we set that =0.
(x-4)(x+3)=0.
This is saying that one quantity times another is 0. So, one or both quantities have to be 0.
So x-4=0 or x+3=0
x=4 or x=-3. These are the points where the y-value is zero.
So (4,0), (-3,0) are the x-intercepts.
The y-intercept is found by setting x=0.
This gives -4*3=-12. So (0,-12) is the y-intercept.
Moreover, the vertex is given by x=-b/(2a) for ax^2+bx+c=0 form. So we have x=1/2 is the vertex. f(1/2)=1/4-1/2-12=-49/4 gives the y value of the vertex.
Lastly, the function will open upward because the leading coefficient of x^2 is positive.
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