How do you graph #y=3x^2+6x-1#?

Answer 1

A quadratic function represents a parabola, hence its vertex and the axis of symmetry can be found from the given equation as follows

y= #3(x^2 +2x)-1# = #3 (x^2 +2x +1) +1-3# = #3(x+1)^2 -2#
Since coefficient of#x^2# is positive, the parabola would open up, its vertex would be at (-1,-2), the axis of symmetry would be line x= -1. X-intercepts would be #-1 +-2/3 sqrt3# The parabola would also cross y axis at point (0, -1). With these inputs the curve can be easily sketched.
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Answer 2

To graph the quadratic equation y = 3x^2 + 6x - 1, you can follow these steps:

  1. Plot the vertex of the parabola, which is located at the point (-b/2a, f(-b/2a)).
  2. Determine the axis of symmetry, which is the vertical line passing through the vertex.
  3. Find at least two additional points on either side of the vertex to sketch the parabola accurately.
  4. Plot these points and draw a smooth curve through them to represent the graph of the equation.
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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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