What is the vertex of #y=3x^2-7x+12#? What are its x-intercepts?

Answer 1

Find vertex of #y = 3x^2 - 7x + 12#.

x-coordinate of vertex: #x = (-b/(2a)) = 7/6# y-coordinate of vertex: #y = y(7/6) = 3(49/36) - 7(7/6) = 12 = 147/36 - 49/6 + 12 =# #= - 147/36 + 432/36 = 285/36 = 7.92# Vertex #(7/6, 7.92)# To find the 2 x-intercepts, solve the quadratic equation: #y = 3x^2 - 7x + 12 = 0.# #D = b^2 - 4ac = 49 - 144 < 0#. There are no x-intercepts. The parabola opens upward and is completely above the x-axis. graph{3x^2 - 7x + 12 [-40, 40, -20, 20]}
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Answer 2

The vertex of the parabola represented by the equation y = 3x^2 - 7x + 12 is (7/6, 59/12). To find the x-intercepts, set y = 0 and solve the quadratic equation 3x^2 - 7x + 12 = 0. The solutions are approximately x ≈ 2.08 and x ≈ 1.45. Therefore, the x-intercepts are approximately (2.08, 0) and (1.45, 0).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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