What are the x-intercepts of the function f(x)=x^2=3x-4? and how do I figure this???

Question

anonymous · Answer

Is that equation supposed to be x^2 + 3x - 4 or maybe x^2 - 3x - 4 ?

anonymous · Answer

If you are working with a parabola, which would be an equation in the form y = f(x) = ax^2 + bx + c and you are looking for x-intercepts, then you set the equation = 0 and use the quadratic formula:$$x = \frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }$$

anonymous · Answer

1) To find the x-intercept of this equation you have to first subtract 3x-4 to get f(x) = x^2-3x+4.

2) Then you set f(x) to 0 and you get 0=x^2-3x+4
3) Find two nubers that add up to -3 and multiply to 4
4) Nothing in this cse works so you have to use the quadratic formula which is $$\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$$

5) a=1, b=-3, and c=4, plug it in and simplify .... you should get $$\frac{ 3\pm i \sqrt{7} }{ 2}$$

6) This shows you that the parabola doesn't touch any of the x-axes because the discriminant is negative and you had to use $$i ^{2} = -1$$

Hope that helps :D