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Find the x-intercept(s) and the coordinates of the vertex for the parabola. Please help me! I'm so very confused.

A) x^2 - 4x + 3; x = 1, (2, 1)
B) 2x^2 - 5x + 3; x = 3, (5, 7)
C) x^2 + 6x + 9; x = -3, (-3, 0)
D) 3x^2 - 2x - 4; x = 2, (2, 4)

1 Answer

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Final answer:

To find x-intercepts, one must set the equation equal to zero and solve for x. The vertex can be found using the formula -b/(2a) for the x-coordinate and substituting it into the equation to find the y-coordinate. The provided solutions need to be checked with these methods to ensure accuracy.

Step-by-step explanation:

The question asks about finding x-intercepts and the coordinates of the vertex for given parabolic equations. To find the x-intercepts, we must solve the equation when y=0 (i.e., set the quadratic equation to zero and solve for x). To find the vertex, we can use the formula -b/(2a) to find the x-coordinate and then plug it back into the original equation to find the y-coordinate.

For example, if we have a quadratic equation of the form ax^2 + bx + c = 0, the x-intercepts are found using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), and the vertex's x-coordinate is found with x = -b/(2a). None of the options provided (A, B, C, D) are correct without further calculations to validate the x-intercepts and vertex coordinates.

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