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How many times does the graph of (f(x)=x^2+4) cut the (x)-axis?

Answer and explanation

Correct answer: (0) times

The direct answer is A: the graph cuts the x-axis 0 times. A point on the x-axis has y=0. For this graph, y=x^2+4. Since \\(x^2\ge0\\) for every real x, we have \\(x^2+4\ge4\\). Thus y can never be 0, so there is no x-intercept. A is correct. Option B would require one real solution of \\(x^2+4=0\\), but this gives \\(x^2=-4\\), impossible for real x. Option C would mean two x-intercepts; although a usual upward quadratic may have two, this one is shifted four units above the x-axis and has none. Option D has no algebraic or graphical basis. The vertex is \\( (0,4)\\), and the entire parabola lies above the x-axis. Memory cue: to count x-axis crossings, set y=0 and count real solutions.

Related tags

Relations-FunctionsStandard-GraphsNo-X-Intercept

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