SOLUTION: Find the number of x-intercepts y=x^2-4x

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Question 735056: Find the number of x-intercepts
y=x^2-4x
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
Hard to say. Try completing the square.
x%5E2-4x%2B4-4=%28x-2%29%5E2-4
Very sure that y has TWO x-intercepts, real ones, since the graph opens upward and has a minimum, vertex at (2, -4).

You could also answer the question by using the general solution to a quadratic equation, or just (probably easier) compute the discriminant.

Discriminant is b%5E2-4ac
which would be for your equation, %28-4%29%5E2-4%2A1%2A0=16. POSITIVE 16. This value for the discriminant, is positive, indicating that y has two real zeros.