Question - Equation of a Parabola with Specific Characteristics
mathaxis of symmetryminimum valueequation of parabolaparabola characteristicsupwards opening parabola
Solution:
The question asks for the equation of a parabola that opens upwards, has a minimum value of 3, and an axis of symmetry at x=3. The general form of a parabola that opens upwards with a vertex at (h, k) is given by:f(x) = a(x - h)² + kwhere "h" is the x-coordinate of the vertex, "k" is the y-coordinate of the vertex (and the minimum value of the parabola, since it opens upwards), and "a" is a positive constant that affects the width of the parabola. Since the parabola opens upwards, a must be positive.Here, we are told the axis of symmetry is x=3, which means h is 3. Also, the parabola has a minimum value of 3, so k is also 3.Therefore, the equation becomes f(x) = a(x - 3)² + 3. The value of "a" is not specified, but any positive value of "a" would suffice for it to open upwards. The simplest form to choose is a=1, to match one of the given options.Looking at the options provided:A. f(x) = (x - 3)² + 3 (This matches our derived equation with a=1, h=3, and k=3)B. f(x) = (x - 3)² - 6 (This parabola also has an axis of symmetry at x=3, but it doesn't have a minimum value of 3, as required)C. f(x) = (x + 3)² - 6 (This one has an axis of symmetry at x=-3, which does not match our requirement)The correct answer is therefore:A. f(x) = (x - 3)² + 3
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