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For what values of a does y=ax^2-8x+4 have no x-intercepts?
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We can find a from the discriminant. The point where the discriminant becomes negative, is where the function no longer has x-intercepts. For ax^2 + bx + c, the discriminant is \[b^2 - 4ac\]so we want \[b^2 - 4ac \ge 0\]
Okay thank you! So, I merely sub the numbers for the letters. then I would get \[a \le 4\] and that would mean any value less or equal to 4. \[(-\infty,4]\] ?????
Exactly! Because if the discriminant (from the quadratic formula) is negative, then the function never touches the x-axis.
Okie dokie. Thank you!!!
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