MHB Can Plugging in (x/2) or (x - 1/2) Determine Real Zeros in a Quadratic Equation?

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Plugging in (x/2) or (x - 1/2) into the function f(x) = x - x^2 can help analyze the real zeros of the quadratic equation. The function y = x - x^2 - 1/2 does not intersect the x-axis, indicating that it has no real zeros. The concavity of f(x) suggests that lowering the graph will not change its ability to cross the x-axis. Evaluating the discriminant, D = b^2 - 4ac, for each quadratic form reveals insights into the nature of the zeros. Understanding these shifts and their effects on the graph is crucial for determining the presence of real zeros.
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I can replace f(x) with x - x^2. Should I plug (x/2) into f(x)? How about (x - 1/2) into f(x)? I need the set up.

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Note the range of $f(x)$. A and B are vertical shifts, C and D are horizontal shifts. Which one of the given shifts would result in the graph of $f(x)$ not crossing the $x-\text{axis}$?
 
Greg said:
Note the range of $f(x)$. A and B are vertical shifts, C and D are horizontal shifts. Which one of the given shifts would result in the graph of $f(x)$ not crossing the $x-\text{axis}$?

Let y = the function.

y = x - x^2 - 1/2 does not cross the line y = 0.
 
Note the concavity of $f(x)$. What does that tell you about lowering the graph of $f(x)$? (the line $y=0$ does not concern us presently).
 
write each quadratic in standard form, $ax^2+bx+c$ ... check each discriminant, $D = b^2-4ac$

you know what the discriminant can tell you about the nature of zeros, right?
 

Thread 'Area of Overlapping Squares'

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