find the intercept of g(n)=-2(3n-1)(2n+1)

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find the intercept of g(n)=-2(3n-1)(2n+1)

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    2021-10-04T12:03:26+00:00

    To find the intercepts of this give function of g(n), we have to find both the points present on the axis. That is, X-Intercept axial or axis point and the Y-Intercept axial or axis point and apply the zero factor principle to get the actual points on the graph for both the respective intercepts. Let me make it simpler, by showing the whole process via the LaTeX interpreter equation editor.

    The X-Intercept is that actual point present in the graphical interpretation where the Y-axis is taken as zero, this makes us to point out the position of X-Intercept points on its X-axis and Y-axis. Take the variable “n” as the variable of “x”, it will not change any context or such, we can take any variables for calculations, it does not hinder the processing of Intercepts for the axial points on a graph.

    \boxed{\mathbf{\therefore \quad -2(3x - 1)(2x + 1) = 0}}

    By the zero factor principle, both of them can be separately calculated as a zero on their either sides of the expression.

    \mathbf{\therefore \quad 3x - 1 = 0}

    \mathbf{3x - 1 + 1 = 0 + 1}

    \mathbf{3x = 1}

    \mathbf{\dfrac{3x}{3} = \dfrac{1}{3}}

    \mathbf{\therefore \quad x = \dfrac{1}{3}}

    Similarly, for the second X-Intercept point for the value of 0 in the Y-axis or Y axial plane in a 2 dimensional Graphical representation is going to be, As per the zero factor principle:

    \mathbf{\therefore \quad 2x + 1 = 0}

    \mathbf{2x + 1 - 1 = 0 - 1}

    \mathbf{2x = - 1}

    \mathbf{\dfrac{2x}{2} = \dfrac{- 1}{2}}

    \mathbf{\therefore \quad x = -\dfrac{1}{2}}

    Then the X-Intercept here becomes with our provided points as:

    \boxed{\mathbf{\underline{X-Intercept: \quad \Bigg(\dfrac{1}{3}, \: 0 \Bigg), \: \: \: \: \Bigg(-\dfrac{1}{2}, \: 0 \Bigg)}}}

    Therefore, for our Y-Intercept axial point the X axial plane will instead turn out to be a value with zero on a Graphical representation to obtain the actual points for Y-axis and the Y-Intercept for x = 0 as a point on the graph itself.

    Just substitute the value of “0” in “x” axis as a variable on the provided expression. Therefore:

    \boxed{\mathbf{= -2(3 \times 0 - 1) (2 \times 0 + 1)}}

    \mathbf{y = - 2 (0 - 1) (0 + 1)}

    \mathbf{y = - 2 (- 1) (0 + 1)}

    \mathbf{y = - 2 (- 1) \times 1}

    \mathbf{y = 2 \times 1 \times 1}

    \mathbf{\therefore \quad y = 2}

    Then, the Y-Intercept would definitely be as per the X-axis lying on the point of zero.

    \boxed{\mathbf{\underline{Y-Intercept: \quad \big(0, \: 2 \big)}}}

    The final coordinating points for X-Intercept and Y-Intercept for their X-axis and Y-axis will be.

    \boxed{\mathbf{\underline{X \: Intercepts: \: \: \Bigg(\dfrac{1}{3}, \: 0 \Bigg), \: \: \: \: \Bigg(-\dfrac{1}{2}, \: 0 \Bigg); \: \: Y \: Intercepts: \: \: \big(0, \: 2 \big)}}}

    Hope it helps.

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